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I have this power series: $$\sum_{k=1}^{\infty} \frac{x^{k+2}}{k+1}.$$ How can I find the sum of this? I have tried different ways to differentiate $\sum_{k=0}^{\infty}x^n=\frac{1}{1-x}$ and get it into the right form, but I can't seem to be getting it right.

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  • $\begingroup$ How about integrating $\frac1{1-x}$ instead? $\endgroup$ – mrtaurho May 27 at 20:35
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Consider $$\sum_{k=1}^\infty x^k = \frac{1}{1-x} - 1$$

Integrating both sides, we get $$\sum_{k=1}^\infty \int_0^x x^k dx = \int_0^x\left(\frac{1}{1-x} - 1\right) dx$$ $$\sum_{k=1}^\infty\frac{x^{k+1}}{x+1} = -\ln(1-x) - x$$

What you need is $$\sum_{k=1}^\infty\frac{x^{k+2}}{k+1} = x\cdot\sum_{k=1}^\infty\frac{x^{k+1}}{x+1} = x(-\ln(1-x) - x)$$

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Hint:

$$f(x)=\sum_{k=1}^\infty\frac{x^{k+2}}{k+1}=x\sum_{k=1}^\infty\frac{x^{k+1}}{k+1}\implies f'(x)=\ldots$$

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For $|x|<1$ $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k}$$ $$\int_{0}^{x}\frac{1}{1-x}dx=\int_{0}^{x}\sum_{k=0}^{\infty} x^{k}dx$$ $$-\log(1-x)=\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}$$ multiply by $x$ $$-x\log(1-x)=\sum_{k=0}^{\infty} \frac{x^{k+2}}{k+1}$$

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$$\sum _{k=1}^\infty x^{k+2}/{(k+1)}=x\sum _{k=1}^\infty x^{k+1}/{(k+1)}$$

Recognize the term $x^{k+1}/{(k+1)}$ as an integral and proceed from there.

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