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

Question

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!!

Answer 1

Note that $$\sum_{k=1}^{n+1}\frac{1}{(n+1)+k}= \sum_{k=1}^{n}\frac{1}{n+k}-\frac1{n+1}+\frac{1}{2n+1}+\frac1{2n+2}$$

$$\begin{align} -\frac1{n+1}+\frac{1}{2n+1}+\frac1{2n+2}&=\frac{1}{2n+1}-\frac{1}{2n+2}\Rightarrow \\\sum_{k=1}^{n+1}\frac{1}{(n+1)+k}&=\sum_{k=1}^{2(n+1)}(-1)^{k+1}\frac{1}{k} \end{align}$$

Answer 2

We have that

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

Answer 3

Hint:

$$\begin{align}\sum_{k=1}^{n}\frac{1}{n+k}&=\frac{1}{n+1}+\sum_{k=2}^{n}\frac{1}{n+k}\\ &=\frac{1}{n+1}+\sum_{k=1}^{n-1}\frac1{n+1+k}\end{align}$$