Radius of convergence for three series
$begingroup$
I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.
1) $sum_{n=0}^{infty}x^{n!}$
2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$
3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$
1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1
Thus the radius of convergence is 1.
On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.
2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.
We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$
3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$
On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.
Are my results correct ? If not, what did I do wrong or forgot ?
Thanks for your help !
real-analysis sequences-and-series convergence power-series divergent-series
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
$endgroup$
add a comment |
$begingroup$
I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.
1) $sum_{n=0}^{infty}x^{n!}$
2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$
3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$
1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1
Thus the radius of convergence is 1.
On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.
2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.
We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$
3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$
On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.
Are my results correct ? If not, what did I do wrong or forgot ?
Thanks for your help !
real-analysis sequences-and-series convergence power-series divergent-series
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
$endgroup$
add a comment |
$begingroup$
I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.
1) $sum_{n=0}^{infty}x^{n!}$
2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$
3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$
1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1
Thus the radius of convergence is 1.
On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.
2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.
We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$
3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$
On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.
Are my results correct ? If not, what did I do wrong or forgot ?
Thanks for your help !
real-analysis sequences-and-series convergence power-series divergent-series
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
$endgroup$
I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.
1) $sum_{n=0}^{infty}x^{n!}$
2) $sum_{n=0}^{infty} frac{1}{1+2^n}x^n$
3) $sum_{n=0}^{infty} frac{1}{1+n3^n}x^n$
1) For the first, $a_n=1$ iff $exists k : k!=n, k in mathbb{N}$, zero elsewhere. So the radius is given by $frac{1}{lim : sup_{n to infty}sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1
Thus the radius of convergence is 1.
On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.
2) Here, we use $rho = frac{a_n}{a_{n+1}}$ So we get : $lim_{n to infty}(frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $frac{2^n}{1+2^n}=frac{2^n}{2^n}frac{1}{frac{1}{2^n}+1}=1$ for $n to infty$, so not toward 0, and thus it diverges.
We also have $frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n to infty$ which isn't the case here, so it also diverges for $x=-2$
3) Here, we get $frac{1+3^{n+1}(n+1)}{1+3^nn}= frac{3^nn}{3^nn}frac{frac{1}{3^nn}+frac{3(n+1)}{n}}{1+ frac{1}{3^nn}}=frac{3}{1}=3$ for $n to infty$
On the border, we have $frac{3^n}{1+n3^n}=frac{1}{frac{1}{3^n}+n}$ and $frac{(-1)^n3^n}{1+n3^n}=frac{(-1)^n}{frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.
Are my results correct ? If not, what did I do wrong or forgot ?
Thanks for your help !
real-analysis sequences-and-series convergence power-series divergent-series
real-analysis sequences-and-series convergence power-series divergent-series
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
edited Dec 20 '18 at 13:04
David G. Stork
12.1k41735
12.1k41735
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
asked Dec 20 '18 at 12:58
PoujhPoujh
6111516
6111516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the first
the ratio test gives the limit
$$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$
if $lvert xrvert<1$, the limit is zero.
if $lvert xrvert>1$ the limit is infinity.
the radius is $R=1$.
When you look for the radius only, you do not need the convergence on the border.
edited Dec 20 '18 at 17:30
user10354138
7,5472925
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
1
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
1
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
|
show 2 more comments
$begingroup$
$(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
So Radius of convergence is $1$
$(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
Radius of convergence is $2$.
$(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$
(When you require radius of convergence, convergence of boundary is not necessary)
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
$endgroup$
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
1
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first
the ratio test gives the limit
$$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$
if $lvert xrvert<1$, the limit is zero.
if $lvert xrvert>1$ the limit is infinity.
the radius is $R=1$.
When you look for the radius only, you do not need the convergence on the border.
edited Dec 20 '18 at 17:30
user10354138
7,5472925
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
1
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
1
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
|
show 2 more comments
$begingroup$
For the first
the ratio test gives the limit
$$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$
if $lvert xrvert<1$, the limit is zero.
if $lvert xrvert>1$ the limit is infinity.
the radius is $R=1$.
When you look for the radius only, you do not need the convergence on the border.
edited Dec 20 '18 at 17:30
user10354138
7,5472925
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
1
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
1
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
|
show 2 more comments
$begingroup$
For the first
the ratio test gives the limit
$$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$
if $lvert xrvert<1$, the limit is zero.
if $lvert xrvert>1$ the limit is infinity.
the radius is $R=1$.
When you look for the radius only, you do not need the convergence on the border.
edited Dec 20 '18 at 17:30
user10354138
7,5472925
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
For the first
the ratio test gives the limit
$$lim_{nto+infty}leftlvert frac{x^{(n+1)!}}{x^{n!}}rightrvert=lim_{nto+infty} lvert xrvert^{ncdot n!}$$
if $lvert xrvert<1$, the limit is zero.
if $lvert xrvert>1$ the limit is infinity.
the radius is $R=1$.
When you look for the radius only, you do not need the convergence on the border.
edited Dec 20 '18 at 17:30
user10354138
7,5472925
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
edited Dec 20 '18 at 17:30
user10354138
7,5472925
edited Dec 20 '18 at 17:30
user10354138
7,5472925
edited Dec 20 '18 at 17:30
user10354138
7,5472925
7,5472925
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 20 '18 at 13:09
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
1
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
1
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
|
show 2 more comments
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
1
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
1
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
Something is wrong with the total rep.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:13
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
What do you mean exactly with "wrong with the total rep." ? Ans just for 1), your method seems in fact a lot easier but would my method also be correct for 1) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:15
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
$begingroup$
@Poujh This post if for managers.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:16
1
1
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
$begingroup$
@Poujh Your answerS are correct.but no need for the border.
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 13:19
1
1
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
$begingroup$
I’m sorry, I don’t follow. How is $frac{x^{(n+1)!}}{x^{n!}} not equal to $frac{x^{n*n!}$
$endgroup$
– R.Jackson
Dec 20 '18 at 13:42
|
show 2 more comments
$begingroup$
$(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
So Radius of convergence is $1$
$(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
Radius of convergence is $2$.
$(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$
(When you require radius of convergence, convergence of boundary is not necessary)
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
$endgroup$
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
1
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
add a comment |
$begingroup$
$(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
So Radius of convergence is $1$
$(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
Radius of convergence is $2$.
$(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$
(When you require radius of convergence, convergence of boundary is not necessary)
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
$endgroup$
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
1
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
add a comment |
$begingroup$
$(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
So Radius of convergence is $1$
$(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
Radius of convergence is $2$.
$(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$
(When you require radius of convergence, convergence of boundary is not necessary)
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
$endgroup$
$(1)$ $$displaystylesum_{n=0}^{infty}x^{n!}<sum_{n=0}^{infty}x^n$$ and $displaystylesum_{n=0}^{infty}x^n$ converges for $|x|<1$.
So Radius of convergence is $1$
$(2)$ Use D'Alembert's ratio test: $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+2^n}{1+2^{n+1}}right||x|<1 implies |x|<2$$
Radius of convergence is $2$.
$(3)$ $$displaystylelim_{ntoinfty}left|dfrac{a_{n+1}x^{n+1}}{a_nx^n}right|<1implieslim_{ntoinfty}left|dfrac{1+n3^n}{1+(n+1)3^{n+1}}right||x|<1 implies |x|<3$$ Radius of convergence is $3$
(When you require radius of convergence, convergence of boundary is not necessary)
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
edited Dec 20 '18 at 13:30
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
answered Dec 20 '18 at 13:10
Yadati KiranYadati Kiran
2,1161622
2,1161622
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
1
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
add a comment |
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
1
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
$begingroup$
Just for 1), do you think my method would also be correct (even if yours is easier) ?
$endgroup$
– Poujh
Dec 20 '18 at 13:19
1
1
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
$begingroup$
@Poujh: Yes it's correct. Your argument for $(1)$ is better than mine.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 13:22
add a comment |
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