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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 $=\dfrac{1}{\sqrt{2}}$

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The radius of convergence is $1$. –  André Nicolas Jun 8 '13 at 15:36
    

2 Answers 2

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 $a_{\sqrt{k}}=k/2^{\sqrt{k}}$.

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

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