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I've been trying to extract the $f(x)$ from the following Taylor series:

$$ \sum_{k=1}^{\infty} (-1)^{k}\frac{kx^{k+1}}{3^{k}} $$

I moved things around a bit to make it look like this:

$$ \sum_{k=1}^{\infty} (-\frac{1}{3})^{k}kx^{k+1} $$

But then I get stuck when I realize the derivative I have to take is:

$$ \frac{d}{dx}x^{k+1} = (k+1)x^{k} $$

As where I need something instead that can take out my $k$. Unless there is a way I can change my $(k)$ value into $(k+1)$, I have no idea what to do next.

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  • $\begingroup$ If it were $kx^{k-1}$ instead of $kx^{k+1}$ in the series, would you know what to do? $\endgroup$ – Daniel Fischer Mar 26 '14 at 19:01
  • $\begingroup$ This series is the derivative of a polylogarithm. $\endgroup$ – Lucian Mar 26 '14 at 20:28
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Hint: $$kx^{k+1}=(k+1)x^{k+1} - x^{k+1}$$ and $$\int (k+1)x^{k+1} dx=\cdots$$

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I ended up figuring it out, I realized that I can just throw as many $x$ variables as I need onto the left of the sigma.

$$ \sum_{k=1}^{inf}(-\frac{1}{3})^{k}kx^{k+1} = |x^{2}|\sum_{k=1}^{inf}(-\frac{1}{3})^{k}kx^{k-1} $$

Where $ \frac{d}{dx}x^{k} = kx^{k-1} $ so...

$$ |x^{2}|\sum_{k=1}^{inf}(-\frac{1}{3})^{k}\frac{d}{dx}x^{k} = |x^{2}|\frac{d}{dx}\sum_{k=1}^{inf}(-\frac{x}{3})^{k}$$

Where that last summation is in geometric series notation, so we can write:

$$ |x^{2}|\frac{d}{dx}\sum_{k=1}^{inf}(-\frac{x}{3})^{k} = |x^{2}|\frac{d}{dx}\frac{-\frac{x}{3}}{1 + \frac{x}{3}} $$

By simplifying and evaluating the derivative, we will get the final correct answer of:

$$ -\frac{3x^{2}}{(x+3)^{2}} $$

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