Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the radius of convergence for the following power series I did some and want to check the answers

$$ \sum\limits_{n=1}^\infty {n^2\over 2^n}x^{n^2} $$

I solved it the following way but I am not sure

$ a_n = \dfrac{n}{2^{n^1/2}} $ if $ n = k^2 $ for some natural number $k$, $ 0 $ Otherwise

Thus $\limsup \sqrt[n]{a_n} = \lim\limits_{n\to\infty}\left( \dfrac{\sqrt[n]{n}}{\sqrt{2}}\right) = \dfrac{1}{\sqrt{2}}$

Radius of convergence $=\dfrac{1}{\sqrt{2}}$

share|cite|improve this question
The radius of convergence is $1$. – André Nicolas Jun 8 '13 at 15:36

Unfortunately $2^{\sqrt{k^2}}\neq 2^{k/2}$, which ruins the calculation. This error was caused by a substitution confusion. If you want $n^2=k$, then $n=\sqrt{k}$ and $a_{\sqrt{k}}=k/2^{\sqrt{k}}$.

share|cite|improve this answer

For convergence we need $C=\lim \sup |{n^2\over 2^n}x^{n^2}|^{1\over n}<1$

share|cite|improve this answer

Your Answer


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.