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I'm having a bit of a problem with finding the function that represents the sum : $$\sum_{n = 1}^\infty\frac{(-1)^{n+1}x^{n+1}}{n(n+1)}$$

I decided to differentiate it a few times and see if I get something that rings a bell and I found that second derivative gives me the sum:
$$\sum_{n=1}^\infty(-x)^{n-1}$$ which is the function $1/1+x$. So I integrated that function twice and found $(x+1)\ln(x+1)-x$.

Now this seems a bit unefficient since it might have worked in this case, but what if it took me 30 differentiations to find a fammiliar sum? So I guess my question is divided into two - first, is the function I found even correct - and second, how would you approach something like this?

Kenny Wong
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dvd280
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    Personally I think your method is very nice! Of course, this won't work in for every example, but then again, most power series don't have a nice description in terms of a well-known function, so we wouldn't expect this method to work all the time! – Kenny Wong May 04 '17 at 08:32
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    This is well done ! For this kind of problems, quite often, differentiation or integration are the ways to explore. – Claude Leibovici May 04 '17 at 08:33
  • To answer your first question, use $ln(1+x) = x - x^2/2 + x^3/3 - ...$. Then expand $(x+1)ln(1+x) - x$ and simplify to get your original summation. – sku May 07 '17 at 00:09

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While I definitely agree that differentiation and integration are among the first things to probe, $$\sum_{n=1}^{\infty}\frac{(-x)^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty}\left(\color{blue}{\frac{1}{n}-\frac{1}{n+1}}\right)(-x)^{n+1}=x\ln(1+x)-\big(x-\ln(1+x)\big)$$ is what I see in cases like this (I mean partial fractions, and "something logarithmic" if lucky).

metamorphy
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