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I need to evaluate the series $\sum_{k=2}^{\infty} k^{-1} x^{3k+1}$ for $|x| < 1$, and then say something about the series for when $x=1$ and $x \neq 1$.

So far I have the following work: we note that $\sum_{k=2}^{\infty} k^{-1} x^{3k+1} = x \sum_{k=2}^{\infty} k^{-1} x^{3k}$. Now consider the series $\sum_{k=2}^{\infty} k^{-1} x^\alpha$. Then \begin{align} \sum_{k=2}^{\infty} k^{-1} x^\alpha = \sum_{k=2}^{\infty} (k^{-1} +k - k) x^{\alpha} = \sum_{k=2}^{\infty} (k^{-1} - k) x^{\alpha} + \sum_{k=2}^{\infty} k x^{\alpha}. \end{align}

This is where I'm stuck. The term $\sum_{k=2}^{\infty} k x^{\alpha}$ is easy to handle, but I can't see what to do with the first term.

Kalypso
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1 Answers1

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Hint:

$$x\sum_{k=2}^{\infty} k^{-1} x^{3k}=x\sum_{k=2}^{\infty} \frac{(x^3)^k}{k}$$

Now, use the Taylor series for $-\ln(1-u)=\frac{u}{1}+\frac{u^2}{2}+\frac{u^3}{3}+\ldots$, in which $u=x^3$.

MrYouMath
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