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How would I find the radius of convergence of the following series?

$$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n}z^{n(n+1)} $$

The ratio test and root test are inconclusive, so I think I have to use the definition of the radius of convergence

$$\frac{1}{R}=\limsup|a_n|^{\frac{1}{n}}.$$

I have been told that the $n$-th coefficient of this series is not $$\frac{(-1)^n}{n}.$$

I am not sure what exactly I should equate to $|a_n|$. Any help is appreciated.

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If $|z|\gt 1$, then $\frac{|z|^{n(n+1)}}n$ does not converge to $0$, and when $|z|\lt 1$, $\sum_n |z|^{n(n+1)} $ is convergent, so the radius of convergence is $1$.

Indeed, the $n$-th coefficient is not $\frac{(-1)^n}n$ as if we write the series into the form $\sum_k b_kx^k$, we have $b_{k(k+1)}=\frac{(-1)^k}k$ and $b_n=0$ if $n$ is not of the form $k(k+1)$ for some integer $k$.

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  • $\begingroup$ Ok so say we want to find out if the series diverges for z = -1, z = i, how would we go about it? $\endgroup$ – John Sweeney Oct 20 '13 at 19:38
  • $\begingroup$ Notice that $n(n+1)$ is an even integer. $\endgroup$ – Davide Giraudo Oct 20 '13 at 19:39

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