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I've been trying to solve this problem for the past few hours and was not able to find a solution to it. The goal is to proof through full induction that the 2 sums are identical.

$$\sum_{\ell=1}^{2n} (-1) ^{\ell+1} \frac{1}{\ell} = \sum_{k=1}^n \frac{1}{n+k}.$$

The furthest I was able to get is:

$$\sum_{\ell=1}^{2(n+1)} (-1)^{\ell+1}\frac{1}{\ell} = \sum_{\ell=1}^{2n} (-1)^{\ell+1}\frac{1}{\ell} + \frac{1}{2n+1}-\frac{1}{2n+2} = \sum_{\ell=1}^{n} \frac{1}{n+l} + \frac{1}{2n+1}-\frac{1}{2n+2}. $$

Have I done anything wrong in my approach up to this point, or am I just to silly to find out how to proceed from here?

amWhy
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On a mission
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  • Which two sums? The Inductive proof should be given to show the equation given holds. To prove my induction, where's the base case for, $n=1$, and state the induction hypothesis. Your question seems to be about the induction hypothesis. – amWhy Oct 26 '21 at 22:05

1 Answers1

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Answer.

I think you gave up on the really last part.

$$\sum_{\ell=1}^{n} \frac{1}{n+l} + \frac{1}{2n+1}-\frac{1}{2n+2}=\frac1{n+1}+\sum_{\ell=1}^{n-1} \frac{1}{n+l+1} + \frac{1}{2n+1}-\frac{1}{2n+2}\\=\sum_{\ell=1}^{n-1} \frac{1}{n+l+1} + \frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{\ell=1}^{n+1} \frac{1}{n+l+1}$$

MH.Lee
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    Thank you very much for the simple and accurate answer. Furthermore yes, it turns out I should've kept going instead of doubting my steps until then. – On a mission Oct 26 '21 at 22:39