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