# generating function for $\sum\limits_k \frac{x^k}{k^2}$?

Does anyone know the generating function $f$ of $$f(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}$$

How can we get it? Thanks!

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If you're asking for a function which has the Taylor series in your question when expanded around $x=0$, the answer is the dilogarithm $\operatorname{Li}_2(x)$.

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If we differentiate we get $$f'(x) = \sum_{k=1}^{\infty} \frac{x^{k-1}}{k}= -\frac{\log(1-x)}{x}$$ so $\displaystyle f(x) = \int^x_0 \frac{-\log(1-t)}{t} dt.$

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The polylogarithm, $\operatorname {Li}_s$ is defined by the infinite series

$$\operatorname {Li}_s(x)=\sum_{k=1}^\infty \frac{x^k}{k^s}$$

Thus your sum is the polylogarithm of order 2. $\operatorname {Li}_2$ is a special case fo the polylogarithm, called the dilogarithm/Spence's function.

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You mean the generating function of the series given by $\tfrac{1}{k^2}$?
Plug "series x^k/k^2" it into wolfram alpha and it tells you it's the polylogarithm for $s=2$.