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Find the $x$ s for which $$\sum_{n=2}^{\infty} \frac{x^{n^2}}{n \log(n)}$$ converges.

How can I do this? My attempt is to write $\sum_{n=2}^{\infty} \frac{x^{n^2}}{n \log(n)}$ in the form $\sum a_n(x-\xi)^n$. How can I do it? Do I have to use the logarithmic function?

  • Use the Root Test, sometimes named after Cauchy. It will take care of everything except $!$ (diverge there, Integral Test) and $-1$ (converge, Alternating Series (Leibniz) Test. – André Nicolas Jun 28 '14 at 19:45
  • You mean this one: $\frac{x^{n^2}}{n \log(n)} \geq \frac{x^n}{n \log(n)}$ ? –  Jun 28 '14 at 19:48
  • No, you need to look up Root Test, we take the $n$-th root of the absolute value, find the limit. The limit of $(n\log n)^{1/n}$ is $1$. so if $x$ has absolute value $\gt 1$, limit does not exist. If $x$ has absolute value $\lt 1$, the limit of the $n$-th root is $0$. If $x=\pm 1$ the limit is $1$, inconclusive. But we can also do comparisons. – André Nicolas Jun 28 '14 at 19:53

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Hint: Use the Root Test, sometimes named after Cauchy. It will take care of everything except $x=1$ (diverge there, Integral Test) and $x=-1$ (converge, Alternating Series (Leibniz) Test.

Instead of the Root Test, you can use the Ratio Test.

If the Root Test is unfamiliar, we can use the fact that $|x|^{n^2}\ge |x|^n$ if $|x|\gt 1$, and $|x|^{n^2}\le |x|^n$ if $|x|\lt 1$ to reduce the problem to a more familiar one.

André Nicolas
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