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I am given the following power series: $$\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt n}(x+1)^{2n+1}$$

I need to find the radius of convergence. Based on a solution I read:

$$ a_n = \begin{cases} \frac {(-1)^k}{\sqrt k} & n=2k+1 \\ 0 & \text{else} \end{cases} = \begin{cases} \frac {-1}{\sqrt k} & n=2k+1 \\ 0 & \text{else} \end{cases}$$

Is this correct? because I don't understand why $(-1)^k=-1 $ here, since k can be even. I would also like to hear your thoughts on how to solve it, Thanks

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    You're right, they confused themselves. – 2017-02-04

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Try this simple trick: denote our series by $\sum\limits_{k=3}^{\infty}a_k(x+1)^k$. Then $a_{2k+1}=\frac{(-1)^k}{\sqrt{k}}$ for $k=1,2,\dots$ and $a_{2k}=0$ for $k=2,3,\dots$. Use Cauchy-Hadamard theorem (with $\limsup$ version) to determine the radius of convergence.