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What is the radius of the convergence of the series?

Please show clearly and help me how to solve this. Thank you!

I know the $R=\dfrac{1}{\limsup|a_n|^{1/n}}$ but I cannot find value of limsup. My actual question to you is this!!?

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Sorry!!! I rewrite the correct form! – UserN48 Feb 23 '13 at 18:28
You still have two useless extra parentheses. – 1015 Feb 23 '13 at 18:30
You can show $\limsup\limits_{n\rightarrow\infty}|a_n|^{1/n}=\limsup\limits_{n\rightarrow \infty}{1\over 5+\cos(n\pi/3) }= $ ?? (what happens for $n$ an odd multiple of $3$?) – David Mitra Feb 23 '13 at 18:46
Yes what is the value of the lımsup? Please can you explicitly write below the way of the solution to find the value? Please! @DavidMitra – UserN48 Feb 23 '13 at 18:53
$limsupn→∞|an|^{1/n}=lim supn→∞{n^{2}(5+cos(nπ/3))}$ What is the value of this? @DavidMitra – UserN48 Feb 23 '13 at 18:57
up vote 1 down vote accepted

So let's try to apply the limsup formula for the radius of convergence, since you say this is your actual question.

First compute $$ \sqrt[n]{|a_n|}=\frac{1}{n^{2/n}(5+\cos(n\pi/3))}. $$

Recall that $\limsup x_n$ is the largest $x$ such that there exists a subsequence of $x_n$ converging to $x$.

Note that $n^{2/n}=\exp (2\log n/n)$ converges to $1$, so that $$ \limsup \sqrt[n]{|a_n|}=\limsup \frac{1}{5+\cos(n\pi/3)}. $$

First oberve that $$ \frac{1}{5+\cos(n\pi/3)}\leq \frac{1}{5-1}=\frac{1}{4} $$ for all $n$. So $\limsup \sqrt[n]{|a_n|}\leq 1/4$.

And now for the extraction $n_k=3(2k+1)$, we have $$ \frac{1}{5+\cos(n_k\pi/3)}=\frac{1}{4}. $$ Hence $\limsup \sqrt[n]{|a_n|}\geq 1/4$.

Finally $\limsup \sqrt[n]{|a_n|}= 1/4$ and by the formula for the radius of convergence, $R=4$.

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Thank you!! @julien – UserN48 Feb 23 '13 at 19:00
You're welcome! It took me some time to get it right... but we finally got it. – 1015 Feb 23 '13 at 19:02

Edit: OP has changed the question, replacing $n^2$ with $n^{56}$, and $5$ with $23$, and $n\pi/3$ with $n\pi/7$. Fundamentally, nothing changes, the radius of convergence is now $22$, same argument.

Original Question: This asked for the radius of convergence of $\displaystyle \sum_1^\infty \frac{x^n}{n^2(5+\sin(n\pi/3))^n}$.

Answer: Applying the Ratio Test, or the Root Test, shows that the radius of convergence is $\ge 4$. You may want to compare with the simpler series $\displaystyle\sum_1^\infty \frac{x^n}{n^24^n}$.

To show that nothing $\gt 4$ will do, show that if $|x|\gt 4$, then the terms do not have limit $0$. The issue is that $\cos(n\pi/3)$ is $-1$ for infinitely many $n$.

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Thanks for helping me realize that I had first gone completely wrong... – 1015 Feb 23 '13 at 19:00

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