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Is there a name for this series?

$$\sum_{k=1}^{\infty}\frac{a^{2k}}{2k}.$$

I know that:

$$\tanh^{-1}(a)=\sum_{k=1}^{\infty}\frac{a^{2k-1}}{2k-1}.$$

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$$-\log(1-z)-\tanh^{-1}(z)=\sum_{k=1}^\infty \frac{z^{2k}}{2k}$$ –  progressiveforest Jul 29 '12 at 5:56
    
Yup I noticed. Was curious, if there is a name. –  J.A Jul 29 '12 at 6:37

1 Answer 1

up vote 2 down vote accepted

I don't believe there's a name for the series you have there, but,

$$\begin{align*} -\log(1-z)&=\sum_{k=1}^\infty\frac{z^k}{k}\\ -\log(1-z^2)&=\sum_{k=1}^\infty\frac{z^{2k}}{k}\\ -\frac12\log(1-z^2)&=\sum_{k=1}^\infty\frac{z^{2k}}{2k}\\ \log\frac1{\sqrt{1-z^2}}&=\sum_{k=1}^\infty\frac{z^{2k}}{2k}\\ \end{align*}$$

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