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I wanted to prove this using induction, but since the RHS is a sum, I can't use the assumption. Therefore I'm stuck in the middle. Is there maybe a way proving that without induction?

If induction is the only way, can you give any hints please? Especially how to get rid of the sum on the RHS?

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

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HINT: You will want to do it by induction. For the induction step, notice that if you replace $n$ by $n+1$, you add

$$(-1)^{2n+2}\frac1{2n+1}+(-1)^{2n+3}\frac1{2n+2}=\frac1{2n+1}-\frac1{2n+2}$$

to the lefthand expression, and on the right you add

$$\sum_{k=1}^{n+1}\frac1{n+1+k}-\sum_{k=1}^n\frac1{n+k}=\sum_{k=2}^{n+2}\frac1{n+k}-\sum_{k=1}^n\frac1{n+k}\;.\tag{1}$$

The difference on the righthand side of $(1)$ allow a lot of telescoping. In fact, you should end up with just three terms.

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  • $\begingroup$ I'm sorry but I don't understand how you got your term on the right hand side (1)? I put $n+1$ in for $n$ but I don't get what you got. $\endgroup$ – Arthur Nov 3 '15 at 5:53
  • $\begingroup$ @Arthur: Do you mean the first summation, $\sum_{k=2}^{n+2}\frac1{n+k}$? $\endgroup$ – Brian M. Scott Nov 3 '15 at 5:55
  • $\begingroup$ I mean the whole line (1). I get $\sum_{k=1}^n\frac{1}{n+1+k}+\frac{1}{2n+2}$. Where does your $-\sum_{k=1}^n\frac{1}{n+k}$ come from? $\endgroup$ – Arthur Nov 3 '15 at 6:00
  • $\begingroup$ @Arthur: I’m not simply evaluating the righthand side of your equation with $n$ replaced by $n+1$: I’m calculating how much that change adds to the righthand side, just as my first calculation shows how much it adds to the lefthand side of your equation. Thus, my first summation in $(1)$ is the righthand side for $n+1$, and I’m subtracting the righthand side for $n$ to get the increase. $\endgroup$ – Brian M. Scott Nov 3 '15 at 6:03
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    $\begingroup$ @Arthur: Looks good! $\endgroup$ – Brian M. Scott Nov 3 '15 at 6:38

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