Radius of convergence of $sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n$












4












$begingroup$


I want to determine the convergence of the following series in dependency of $x$:



$sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n=frac{3}{4}x+frac{2}{5}x^2+frac{1}{4}x^3+frac{3}{17}x^4+ ... $



How can I solve this?



EDIT:
@Winther said, I should try the ratio test:



$q := lim_{n to infty} left| frac{a_{n+1}}{a_n}right| $



So we get



$q = lim_{n to infty} left| frac{frac{n+3}{2(n+1)^2+2}}{frac{n+2}{2n^2+2}}right| = lim_{n to infty} left| frac{n+3}{2(n+1)^2+2}frac{2n^2+2}{n+2} right| = lim_{n to infty} left| frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}right| = lim_{n to infty} left| frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}right| $



With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:



$ q = lim_{n to infty} left| frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}right|= 1$



So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).



Should I try another test (e. g. the root test)?



EDIT 2: Corrected the first coefficients.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you try the ratio test? It's the simplest test to apply and it works
    $endgroup$
    – Winther
    Dec 22 '18 at 20:58








  • 3




    $begingroup$
    The first numerical coefficients do not correspond to the general formula.
    $endgroup$
    – Bernard
    Dec 22 '18 at 21:03






  • 2




    $begingroup$
    @Joe After you have expanded everything out, the standard trick is then to divide both the numerator and the denominator by some power of $n$. As Justin has mentioned, we have a cubic over a cubic. Therefore, the natural thing to do is to divide both the numerator and the denominator by $n^3$. Do you know how to proceed?
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 21:56








  • 4




    $begingroup$
    +1 for showing work after getting hints.
    $endgroup$
    – zwim
    Dec 22 '18 at 22:27






  • 2




    $begingroup$
    @Joe The question asks for which $x$ does the power series converges. What you have shown is that the radius of convergence for this power series is $1$, which means that this power series converges on the open interval of radius $1$ centered at $0$ (i.e. when $x in (-1, 1)$) and diverges away from it (i.e. when $x in (-infty, -1) cup (1, +infty)$. However, you still need to find out if this power series converges when $x = -1$ or $1$.
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 22:53


















4












$begingroup$


I want to determine the convergence of the following series in dependency of $x$:



$sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n=frac{3}{4}x+frac{2}{5}x^2+frac{1}{4}x^3+frac{3}{17}x^4+ ... $



How can I solve this?



EDIT:
@Winther said, I should try the ratio test:



$q := lim_{n to infty} left| frac{a_{n+1}}{a_n}right| $



So we get



$q = lim_{n to infty} left| frac{frac{n+3}{2(n+1)^2+2}}{frac{n+2}{2n^2+2}}right| = lim_{n to infty} left| frac{n+3}{2(n+1)^2+2}frac{2n^2+2}{n+2} right| = lim_{n to infty} left| frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}right| = lim_{n to infty} left| frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}right| $



With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:



$ q = lim_{n to infty} left| frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}right|= 1$



So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).



Should I try another test (e. g. the root test)?



EDIT 2: Corrected the first coefficients.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you try the ratio test? It's the simplest test to apply and it works
    $endgroup$
    – Winther
    Dec 22 '18 at 20:58








  • 3




    $begingroup$
    The first numerical coefficients do not correspond to the general formula.
    $endgroup$
    – Bernard
    Dec 22 '18 at 21:03






  • 2




    $begingroup$
    @Joe After you have expanded everything out, the standard trick is then to divide both the numerator and the denominator by some power of $n$. As Justin has mentioned, we have a cubic over a cubic. Therefore, the natural thing to do is to divide both the numerator and the denominator by $n^3$. Do you know how to proceed?
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 21:56








  • 4




    $begingroup$
    +1 for showing work after getting hints.
    $endgroup$
    – zwim
    Dec 22 '18 at 22:27






  • 2




    $begingroup$
    @Joe The question asks for which $x$ does the power series converges. What you have shown is that the radius of convergence for this power series is $1$, which means that this power series converges on the open interval of radius $1$ centered at $0$ (i.e. when $x in (-1, 1)$) and diverges away from it (i.e. when $x in (-infty, -1) cup (1, +infty)$. However, you still need to find out if this power series converges when $x = -1$ or $1$.
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 22:53
















4












4








4


2



$begingroup$


I want to determine the convergence of the following series in dependency of $x$:



$sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n=frac{3}{4}x+frac{2}{5}x^2+frac{1}{4}x^3+frac{3}{17}x^4+ ... $



How can I solve this?



EDIT:
@Winther said, I should try the ratio test:



$q := lim_{n to infty} left| frac{a_{n+1}}{a_n}right| $



So we get



$q = lim_{n to infty} left| frac{frac{n+3}{2(n+1)^2+2}}{frac{n+2}{2n^2+2}}right| = lim_{n to infty} left| frac{n+3}{2(n+1)^2+2}frac{2n^2+2}{n+2} right| = lim_{n to infty} left| frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}right| = lim_{n to infty} left| frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}right| $



With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:



$ q = lim_{n to infty} left| frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}right|= 1$



So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).



Should I try another test (e. g. the root test)?



EDIT 2: Corrected the first coefficients.






sequences-and-series convergence







share|cite|improve this question











$endgroup$




I want to determine the convergence of the following series in dependency of $x$:



$sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n=frac{3}{4}x+frac{2}{5}x^2+frac{1}{4}x^3+frac{3}{17}x^4+ ... $



How can I solve this?



EDIT:
@Winther said, I should try the ratio test:



$q := lim_{n to infty} left| frac{a_{n+1}}{a_n}right| $



So we get



$q = lim_{n to infty} left| frac{frac{n+3}{2(n+1)^2+2}}{frac{n+2}{2n^2+2}}right| = lim_{n to infty} left| frac{n+3}{2(n+1)^2+2}frac{2n^2+2}{n+2} right| = lim_{n to infty} left| frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}right| = lim_{n to infty} left| frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}right| $



With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:



$ q = lim_{n to infty} left| frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}right|= 1$



So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).



Should I try another test (e. g. the root test)?



EDIT 2: Corrected the first coefficients.






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 22:34







Joe

















asked Dec 22 '18 at 20:55









JoeJoe

385




385








  • 1




    $begingroup$
    Did you try the ratio test? It's the simplest test to apply and it works
    $endgroup$
    – Winther
    Dec 22 '18 at 20:58








  • 3




    $begingroup$
    The first numerical coefficients do not correspond to the general formula.
    $endgroup$
    – Bernard
    Dec 22 '18 at 21:03






  • 2




    $begingroup$
    @Joe After you have expanded everything out, the standard trick is then to divide both the numerator and the denominator by some power of $n$. As Justin has mentioned, we have a cubic over a cubic. Therefore, the natural thing to do is to divide both the numerator and the denominator by $n^3$. Do you know how to proceed?
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 21:56








  • 4




    $begingroup$
    +1 for showing work after getting hints.
    $endgroup$
    – zwim
    Dec 22 '18 at 22:27






  • 2




    $begingroup$
    @Joe The question asks for which $x$ does the power series converges. What you have shown is that the radius of convergence for this power series is $1$, which means that this power series converges on the open interval of radius $1$ centered at $0$ (i.e. when $x in (-1, 1)$) and diverges away from it (i.e. when $x in (-infty, -1) cup (1, +infty)$. However, you still need to find out if this power series converges when $x = -1$ or $1$.
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 22:53
















  • 1




    $begingroup$
    Did you try the ratio test? It's the simplest test to apply and it works
    $endgroup$
    – Winther
    Dec 22 '18 at 20:58








  • 3




    $begingroup$
    The first numerical coefficients do not correspond to the general formula.
    $endgroup$
    – Bernard
    Dec 22 '18 at 21:03






  • 2




    $begingroup$
    @Joe After you have expanded everything out, the standard trick is then to divide both the numerator and the denominator by some power of $n$. As Justin has mentioned, we have a cubic over a cubic. Therefore, the natural thing to do is to divide both the numerator and the denominator by $n^3$. Do you know how to proceed?
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 21:56








  • 4




    $begingroup$
    +1 for showing work after getting hints.
    $endgroup$
    – zwim
    Dec 22 '18 at 22:27






  • 2




    $begingroup$
    @Joe The question asks for which $x$ does the power series converges. What you have shown is that the radius of convergence for this power series is $1$, which means that this power series converges on the open interval of radius $1$ centered at $0$ (i.e. when $x in (-1, 1)$) and diverges away from it (i.e. when $x in (-infty, -1) cup (1, +infty)$. However, you still need to find out if this power series converges when $x = -1$ or $1$.
    $endgroup$
    – Alex Vong
    Dec 22 '18 at 22:53










1




1




$begingroup$
Did you try the ratio test? It's the simplest test to apply and it works
$endgroup$
– Winther
Dec 22 '18 at 20:58






$begingroup$
Did you try the ratio test? It's the simplest test to apply and it works
$endgroup$
– Winther
Dec 22 '18 at 20:58






3




3




$begingroup$
The first numerical coefficients do not correspond to the general formula.
$endgroup$
– Bernard
Dec 22 '18 at 21:03




$begingroup$
The first numerical coefficients do not correspond to the general formula.
$endgroup$
– Bernard
Dec 22 '18 at 21:03




2




2




$begingroup$
@Joe After you have expanded everything out, the standard trick is then to divide both the numerator and the denominator by some power of $n$. As Justin has mentioned, we have a cubic over a cubic. Therefore, the natural thing to do is to divide both the numerator and the denominator by $n^3$. Do you know how to proceed?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:56






$begingroup$
@Joe After you have expanded everything out, the standard trick is then to divide both the numerator and the denominator by some power of $n$. As Justin has mentioned, we have a cubic over a cubic. Therefore, the natural thing to do is to divide both the numerator and the denominator by $n^3$. Do you know how to proceed?
$endgroup$
– Alex Vong
Dec 22 '18 at 21:56






4




4




$begingroup$
+1 for showing work after getting hints.
$endgroup$
– zwim
Dec 22 '18 at 22:27




$begingroup$
+1 for showing work after getting hints.
$endgroup$
– zwim
Dec 22 '18 at 22:27




2




2




$begingroup$
@Joe The question asks for which $x$ does the power series converges. What you have shown is that the radius of convergence for this power series is $1$, which means that this power series converges on the open interval of radius $1$ centered at $0$ (i.e. when $x in (-1, 1)$) and diverges away from it (i.e. when $x in (-infty, -1) cup (1, +infty)$. However, you still need to find out if this power series converges when $x = -1$ or $1$.
$endgroup$
– Alex Vong
Dec 22 '18 at 22:53






$begingroup$
@Joe The question asks for which $x$ does the power series converges. What you have shown is that the radius of convergence for this power series is $1$, which means that this power series converges on the open interval of radius $1$ centered at $0$ (i.e. when $x in (-1, 1)$) and diverges away from it (i.e. when $x in (-infty, -1) cup (1, +infty)$. However, you still need to find out if this power series converges when $x = -1$ or $1$.
$endgroup$
– Alex Vong
Dec 22 '18 at 22:53












2 Answers
2






active

oldest

votes


















6












$begingroup$

Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
to 1$).



Therefore
$f(x)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$sumlimits_{n=1}^{infty} x^n$
which is
$-1 < x < -1$.



At $x=1$,
the series is
$f(1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2}
$
which diverges like the
harmonic series.



At $x=-1$,
the series is
$f(-1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} (-1)^n
$
which converges
because it is
an alternating series
with decreasing terms.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
    $endgroup$
    – JavaMan
    Dec 23 '18 at 4:18










  • $begingroup$
    @JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
    $endgroup$
    – Joe
    Dec 23 '18 at 12:25










  • $begingroup$
    @Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
    $endgroup$
    – marty cohen
    Dec 23 '18 at 15:04










  • $begingroup$
    @Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
    $endgroup$
    – JavaMan
    Dec 26 '18 at 3:22



















2












$begingroup$

If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $xne0$,
$$
lim_{ntoinfty}
left|,frac{dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{dfrac{n+2}{2n^2+2}x^n},right|
=lim_{ntoinfty}frac{n+3}{n+2}frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.



Thus the radius of convergence is $1$.



There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
frac{1}{R}=limsup_{ntoinfty}sqrt[n]{a_n}
$$
but this can be difficult to compute.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It should be 1/R in Hadamard’s
    $endgroup$
    – Abrb
    Dec 23 '18 at 1:50












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049843%2fradius-of-convergence-of-sum-limits-n-1-infty-fracn22n22-xn%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
to 1$).



Therefore
$f(x)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$sumlimits_{n=1}^{infty} x^n$
which is
$-1 < x < -1$.



At $x=1$,
the series is
$f(1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2}
$
which diverges like the
harmonic series.



At $x=-1$,
the series is
$f(-1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} (-1)^n
$
which converges
because it is
an alternating series
with decreasing terms.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
    $endgroup$
    – JavaMan
    Dec 23 '18 at 4:18










  • $begingroup$
    @JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
    $endgroup$
    – Joe
    Dec 23 '18 at 12:25










  • $begingroup$
    @Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
    $endgroup$
    – marty cohen
    Dec 23 '18 at 15:04










  • $begingroup$
    @Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
    $endgroup$
    – JavaMan
    Dec 26 '18 at 3:22
















6












$begingroup$

Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
to 1$).



Therefore
$f(x)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$sumlimits_{n=1}^{infty} x^n$
which is
$-1 < x < -1$.



At $x=1$,
the series is
$f(1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2}
$
which diverges like the
harmonic series.



At $x=-1$,
the series is
$f(-1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} (-1)^n
$
which converges
because it is
an alternating series
with decreasing terms.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
    $endgroup$
    – JavaMan
    Dec 23 '18 at 4:18










  • $begingroup$
    @JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
    $endgroup$
    – Joe
    Dec 23 '18 at 12:25










  • $begingroup$
    @Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
    $endgroup$
    – marty cohen
    Dec 23 '18 at 15:04










  • $begingroup$
    @Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
    $endgroup$
    – JavaMan
    Dec 26 '18 at 3:22














6












6








6





$begingroup$

Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
to 1$).



Therefore
$f(x)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$sumlimits_{n=1}^{infty} x^n$
which is
$-1 < x < -1$.



At $x=1$,
the series is
$f(1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2}
$
which diverges like the
harmonic series.



At $x=-1$,
the series is
$f(-1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} (-1)^n
$
which converges
because it is
an alternating series
with decreasing terms.






share|cite|improve this answer









$endgroup$



Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
to 1$).



Therefore
$f(x)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$sumlimits_{n=1}^{infty} x^n$
which is
$-1 < x < -1$.



At $x=1$,
the series is
$f(1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2}
$
which diverges like the
harmonic series.



At $x=-1$,
the series is
$f(-1)
=sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} (-1)^n
$
which converges
because it is
an alternating series
with decreasing terms.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 0:46









marty cohenmarty cohen

75.8k549130




75.8k549130












  • $begingroup$
    Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
    $endgroup$
    – JavaMan
    Dec 23 '18 at 4:18










  • $begingroup$
    @JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
    $endgroup$
    – Joe
    Dec 23 '18 at 12:25










  • $begingroup$
    @Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
    $endgroup$
    – marty cohen
    Dec 23 '18 at 15:04










  • $begingroup$
    @Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
    $endgroup$
    – JavaMan
    Dec 26 '18 at 3:22


















  • $begingroup$
    Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
    $endgroup$
    – JavaMan
    Dec 23 '18 at 4:18










  • $begingroup$
    @JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
    $endgroup$
    – Joe
    Dec 23 '18 at 12:25










  • $begingroup$
    @Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
    $endgroup$
    – marty cohen
    Dec 23 '18 at 15:04










  • $begingroup$
    @Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
    $endgroup$
    – JavaMan
    Dec 26 '18 at 3:22
















$begingroup$
Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
$endgroup$
– JavaMan
Dec 23 '18 at 4:18




$begingroup$
Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
$endgroup$
– JavaMan
Dec 23 '18 at 4:18












$begingroup$
@JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
$endgroup$
– Joe
Dec 23 '18 at 12:25




$begingroup$
@JavaMan Can you explain the first line to me? I understand, that the first part is convergent against 1 and if $left | x right | < 1$ the series converges. Why does the series diverge at $x=1$ and converge at $x = -1$ (Is this correct, or is there something switched?)
$endgroup$
– Joe
Dec 23 '18 at 12:25












$begingroup$
@Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
$endgroup$
– marty cohen
Dec 23 '18 at 15:04




$begingroup$
@Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
$endgroup$
– marty cohen
Dec 23 '18 at 15:04












$begingroup$
@Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
$endgroup$
– JavaMan
Dec 26 '18 at 3:22




$begingroup$
@Joe The point is that if $f(n)$ an $g(n)$ are nonzero polynomials (or even nonzero powers of $n$), then $lim_{nto infty}(f(n)/g(n))^{1/n}$, so that the ratio and root tests essentially ignore these functions when being implemented. Therefore, when determine the interval of convergence, you can ignore them (except at the endpoints).
$endgroup$
– JavaMan
Dec 26 '18 at 3:22











2












$begingroup$

If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $xne0$,
$$
lim_{ntoinfty}
left|,frac{dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{dfrac{n+2}{2n^2+2}x^n},right|
=lim_{ntoinfty}frac{n+3}{n+2}frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.



Thus the radius of convergence is $1$.



There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
frac{1}{R}=limsup_{ntoinfty}sqrt[n]{a_n}
$$
but this can be difficult to compute.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It should be 1/R in Hadamard’s
    $endgroup$
    – Abrb
    Dec 23 '18 at 1:50
















2












$begingroup$

If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $xne0$,
$$
lim_{ntoinfty}
left|,frac{dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{dfrac{n+2}{2n^2+2}x^n},right|
=lim_{ntoinfty}frac{n+3}{n+2}frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.



Thus the radius of convergence is $1$.



There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
frac{1}{R}=limsup_{ntoinfty}sqrt[n]{a_n}
$$
but this can be difficult to compute.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It should be 1/R in Hadamard’s
    $endgroup$
    – Abrb
    Dec 23 '18 at 1:50














2












2








2





$begingroup$

If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $xne0$,
$$
lim_{ntoinfty}
left|,frac{dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{dfrac{n+2}{2n^2+2}x^n},right|
=lim_{ntoinfty}frac{n+3}{n+2}frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.



Thus the radius of convergence is $1$.



There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
frac{1}{R}=limsup_{ntoinfty}sqrt[n]{a_n}
$$
but this can be difficult to compute.






share|cite|improve this answer











$endgroup$



If the ratio test works, there's no need to check with other tests; in your case you want to compute, for $xne0$,
$$
lim_{ntoinfty}
left|,frac{dfrac{(n+1)+2}{2(n+1)^2+2}x^{n+1}}{dfrac{n+2}{2n^2+2}x^n},right|
=lim_{ntoinfty}frac{n+3}{n+2}frac{2(n+1)^2+2}{2n^2+2}|x|=|x|
$$
This limit is $<1$ if and only if $|x|<1$.



Thus the radius of convergence is $1$.



There are cases where the limit for the ratio test doesn't exist; other tests should be used. The “universal” test is Hadamard's:
$$
frac{1}{R}=limsup_{ntoinfty}sqrt[n]{a_n}
$$
but this can be difficult to compute.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 23 '18 at 10:00

























answered Dec 22 '18 at 23:38









egregegreg

186k1486209




186k1486209












  • $begingroup$
    It should be 1/R in Hadamard’s
    $endgroup$
    – Abrb
    Dec 23 '18 at 1:50


















  • $begingroup$
    It should be 1/R in Hadamard’s
    $endgroup$
    – Abrb
    Dec 23 '18 at 1:50
















$begingroup$
It should be 1/R in Hadamard’s
$endgroup$
– Abrb
Dec 23 '18 at 1:50




$begingroup$
It should be 1/R in Hadamard’s
$endgroup$
– Abrb
Dec 23 '18 at 1:50


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049843%2fradius-of-convergence-of-sum-limits-n-1-infty-fracn22n22-xn%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
Read more