Prove that $\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$ I am trying to prove that $$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$$ for $n \in \mathbb{N}$.
My approach is to prove this by induction and this is what I got so far:
$$\sum_{k=1}^{2(n+1)} \frac{(-1)^{k-1}}{k} = (\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}) + \frac{(-1)^{(2n+1)-1}}{2n+1} + \frac{(-1)^{(2n+2)-1}}{2n+2} = (\sum_{k=1}^n \frac{1}{n+k}) + \frac{(-1)^{(2n+1)-1}}{2n+1} + \frac{(-1)^{(2n+2)-1}}{2n+2}$$
However, I am not aware how to proceed from here to get to: $$\sum_{k=1}^{n+1} \frac{1}{(n+1)+k}?$$
I would greatly appreciate any advice/solutions!

Got it, working my way backwards using Maximilian Janisch advice:
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k}=\sum_{k=2}^{n+2} \frac{1}{n+k} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} - \frac{1}{2n+2} = (\sum_{k=1}^n \frac{1}{n+k}) + \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2}$$
Thanks to everyone who helped!
