Prove that $\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n} \frac{1}{k}(-1)^{k-1}$ using induction I'm trying to prove (using induction) that:
$$\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n} \frac{1}{k}(-1)^{k-1}.$$
I have found problems when I tried to establish an induction hypothesis and solving this because I've learned to do things like:
$$ \sum_{k=1}^{n+1} \frac{1}{k}= \sum_{k=1}^{n}  \frac{1}{k} + \frac{1}{n+1}.$$
But, in this case, $n$ appears in both parts of summation and I have no idea  how make a relation with  
$$\sum_{k=1}^n \frac{1}{n+k}  $$ and $$\sum_{k=1}^{n+1} \frac{1}{n+1+k}.  $$ 
Because I've seen tha, the case with "$n+1$" should be like: 
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k} = \sum_{k=1}^{2n+2} \frac{1}{k}(-1)^{k-1}$$
and I cant find a connection between 
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k} $$ and $$\sum_{k=1}^{n} \frac{1}{n+k}.$$
Could anyone help me with this?

We can write
$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{n+1+k}&=\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+2}\\&=\left(-\frac{1}{n+1}+\frac{1}{n+1}\right)+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+2}\\&=\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\&=\left(\sum_{k=1}^{n}\frac{1}{n+k}\right)-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\end{align}$$
