Proof $\sum_{k=1}^{2n} (-1)^{1+k}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k}$ by induction. I'm trying to show the following formula:
$$\sum_{k=1}^{2n} (-1)^{1+k}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k}$$
I have already verified the formula with $n=1$, now I continue with the induction:
$$ \sum_{k=1}^{2(n+1)}(-1)^{k+1}\frac{1}{k} = \sum_{k=1}^{2n}(-1)^{k+1}\frac{1}{k}+\frac{1}{2n+1}-\frac{1}{2n+2} \\
\rightarrow\sum_{k=1}^{2(n+1)}(-1)^{k+1}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k} +\frac{1}{2n+1}-\frac{1}{2n+2}$$
But now I have no idea how to get $\sum_{k=1}^{n+1}\frac{1}{(n+1)+k}$, I have been trying a lot of algebra or changing the index but I don't see the "magic" step. Any idea is welcome!!

We have that
$$\dots= \sum_{k=1}^{n}\frac{1}{n+k} +\frac{1}{2n+1}-\frac{1}{2n+2}=\\=\left(\frac1{n+1}+\sum_{k=1}^{n+1}\frac{1}{(n+1)+k} -\frac{1}{2n+1}-\frac{1}{2n+2}\right)+\frac{1}{2n+1}-\frac{1}{2n+2}=\\=\sum_{k=1}^{n+1}\frac{1}{(n+1)+k}$$
