Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}{(5+\cos(n\pi/3))^{n}}}$$

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!!?

share|improve this question
    
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

2 Answers 2

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, $R=4$.

share|improve this answer
    
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 $n$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.