Radius of convergence of power series $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$
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Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.
My try:
Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get
$displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or
the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??
sequences-and-series proof-verification power-series alternative-proof
asked yesterday
Yadati Kiran
467
add a comment |
up vote
2
down vote
favorite
Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.
My try:
Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get
$displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or
the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??
sequences-and-series proof-verification power-series alternative-proof
asked yesterday
Yadati Kiran
467
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.
My try:
Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get
$displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or
the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??
sequences-and-series proof-verification power-series alternative-proof
asked yesterday
Yadati Kiran
467
Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.
My try:
Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get
$displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or
the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??
sequences-and-series proof-verification power-series alternative-proof
sequences-and-series proof-verification power-series alternative-proof
asked yesterday
Yadati Kiran
467
asked yesterday
Yadati Kiran
467
asked yesterday
Yadati Kiran
467
asked yesterday
Yadati Kiran
467
asked yesterday
Yadati Kiran
467
467
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.
answered yesterday
Fred
41.5k1641
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.
answered yesterday
Fred
41.5k1641
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
add a comment |
up vote
1
down vote
accepted
That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.
answered yesterday
Fred
41.5k1641
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.
answered yesterday
Fred
41.5k1641
That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.
answered yesterday
Fred
41.5k1641
answered yesterday
Fred
41.5k1641
answered yesterday
Fred
41.5k1641
answered yesterday
Fred
41.5k1641
41.5k1641
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
add a comment |
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
ok yeah ! I calculated the length of the interval. Thanks @Fred.
– Yadati Kiran
yesterday
add a comment |
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