Radius of convergence of $sumlimits_{n=1}^{infty} frac{n+2}{2n^2+2} x^n$
$begingroup$
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
asked Dec 22 '18 at 20:55
JoeJoe
385
$endgroup$
|
show 7 more comments
$begingroup$
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
asked Dec 22 '18 at 20:55
JoeJoe
385
$endgroup$
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
|
show 7 more comments
$begingroup$
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
asked Dec 22 '18 at 20:55
JoeJoe
385
$endgroup$
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
asked Dec 22 '18 at 20:55
JoeJoe
385
asked Dec 22 '18 at 20:55
JoeJoe
385
edited Dec 22 '18 at 22:34
Joe
asked Dec 22 '18 at 20:55
JoeJoe
385
asked Dec 22 '18 at 20:55
JoeJoe
385
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
|
show 7 more comments
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
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
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$begingroup$
It should be 1/R in Hadamard’s
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– Abrb
Dec 23 '18 at 1:50
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
$endgroup$
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.
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
answered Dec 23 '18 at 0:46
marty cohenmarty cohen
75.8k549130
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
add a comment |
$begingroup$
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
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@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?)
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– Joe
Dec 23 '18 at 12:25
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@Joe: I explained why the series converges at x=-1 and divergences at x=1. What didn't you understand?
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– marty cohen
Dec 23 '18 at 15:04
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@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).
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– JavaMan
Dec 26 '18 at 3:22
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Your first line is always one I have trouble getting across to my calc students. It's put very succinctly here!
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– 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
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
$endgroup$
$begingroup$
It should be 1/R in Hadamard’s
$endgroup$
– Abrb
Dec 23 '18 at 1:50
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
$endgroup$
$begingroup$
It should be 1/R in Hadamard’s
$endgroup$
– Abrb
Dec 23 '18 at 1:50
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
$endgroup$
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.
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
edited Dec 23 '18 at 10:00
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
answered Dec 22 '18 at 23:38
egregegreg
186k1486209
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
add a comment |
$begingroup$
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
add a comment |
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1
$begingroup$
Did you try the ratio test? It's the simplest test to apply and it works
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– Winther
Dec 22 '18 at 20:58
3
$begingroup$
The first numerical coefficients do not correspond to the general formula.
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– 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?
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– Alex Vong
Dec 22 '18 at 21:56
4
$begingroup$
+1 for showing work after getting hints.
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– 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