Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following: 
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Starting with the base case $N = 1$:
$$\displaystyle \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} = \frac{1}{1} + \frac{-1}{2} = \frac{1}{2} = \frac{1}{1+1} = \sum^{1}_{n=1} \frac{1}{N+n}.$$
My problem is the inductive step for $N+1$, starting with
$$\displaystyle \sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N+1}_{n=1} \frac{1}{(N+1)+n}.$$
And now my problem:
\begin{align}
\sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} &\Leftrightarrow \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} \\[1em]
&\underset{\Leftrightarrow}{\text{ind. hyp.}} \sum^{N}_{n=1} \frac{1}{N + n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n}
\end{align}
Is this the correct start, and if so, how do I continue?

It's easy to prove it using integration,
$$\sum_{k=1}^r\frac{1}{k+r}=\int_0^1\sum_{k=1}^r x^{k+r-1}\ dx=\int_0^1\frac{x^r-x^{2r}}{1-x}\ dx$$
$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag1$$
$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2r}(-1)^{k-1}\int_0^1 x^{k-1}\ dx=\int_0^1\sum_{k=1}^{2r}(-x)^{k-1}\ dx$$
$$=\int_0^1\frac{1-x^{2r}}{1+x}\ dx=\ln2-\int_0^1\frac{x^{2r}}{1+x}\ dx=H_{2r}-H_r\tag2$$
where the last result, follows from $\int_0^1\frac{x^{2r}}{1+x}dx=\ln2+H_r-H_{2r}$.
Hence by $(1)$ and $(2)$ , we conclude that
$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}$$
