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?
 A: Claim: For all $n\geq 1$, the statement
$$
P(n) : \sum_{i=1}^{2n}\frac{(-1)^{i-1}}{i}=\sum_{i=1}^n\frac{1}{n+i}
$$
is true.

Note: The key to giving an inductive proof here lies in understanding how to reindex a sum; that is, the main obstacle you face in using the inductive hypothesis is realizing how to coax $\sum_{i=1}^k\frac{1}{k+i}$ out of $\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}$. To that end, first observe the following:
$$
\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}=\sum_{i=2}^{k+2}\frac{1}{k+i}=\sum_{i=1}^k\frac{1}{k+i}-\frac{1}{k+1}+\frac{1}{k+(k+2)}+\frac{1}{k+(k+1)}.\tag{$\dagger$}
$$
Now observe that $(\dagger)$ may be rewritten as
$$
\sum_{i=1}^k\frac{1}{k+i}=\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}+\frac{1}{k+1}-\frac{1}{2k+2}-\frac{1}{2k+1}.\tag{$\dagger\!\dagger$}
$$
With all of that in mind, we can prove the original claim.

Proof. For the base step, we must confirm that $P(1)$ is true, something you have already done. Thus, the base case checks out. 
Inductive step $P(k)\to P(k+1)$: Fix some $k\geq 1$. Assume that
$$
P(k) : \color{green}{\sum_{i=1}^{2k}\frac{(-1)^{i-1}}{i}=\sum_{i=1}^{k}\frac{1}{k+i}}
$$
holds. To be proved is that
$$
P(k+1) : \color{blue}{\underbrace{\sum_{i=1}^{2(k+1)}\frac{(-1)^{i-1}}{i}}_{\text{LHS}}=\underbrace{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}_{\text{RHS}}}
$$
follows. Beginning with the left side of $P(k+1)$,
\begin{align}
\text{LHS} &= \color{blue}{\sum_{i=1}^{2(k+1)}\frac{(-1)^{i-1}}{i}}\tag{definition}\\[1em]
&= \color{green}{\sum_{i=1}^{2k}\frac{(-1)^{i-1}}{i}}-\frac{1}{2k+2}+\frac{1}{2k+1}\tag{by defn. of $\Sigma$}\\[1em]
&= \color{green}{\sum_{i=1}^k\frac{1}{k+i}}-\frac{1}{2k+2}+\frac{1}{2k+1}\tag{by $P(k)$}\\[1em]
&= \left(\color{blue}{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}+\frac{1}{k+1}-\frac{1}{2k+2}-\frac{1}{2k+1}\right)-\frac{1}{2k+2}+\frac{1}{2k+1}\tag{$\dagger\!\dagger$}\\[1em]
&= \color{blue}{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}+\frac{1}{k+1}-\frac{2}{2k+2}\tag{simplify}\\[1em]
&= \color{blue}{\sum_{i=1}^{k+1}\frac{1}{(k+1)+i}}\tag{$k\neq -1,-1/2$}\\[1em]
&= \text{RHS}\tag{definition}
\end{align}
one arrives at the right side of $P(k+1)$, thereby showing $P(k+1)$ is also true, completing the inductive step.
Conclusion. By mathematical induction, it is proved that for $n\geq 1$ the statement $P(1)$ is true. $\blacksquare$ 
A: Instead of induction, you could start with
$\displaystyle \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\sum_{n=1}^{2N}\frac{1}{n}-2\sum_{k=1}^N\frac{1}{2k}=\sum_{n=1}^{2N}\frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n}$
A: you may find it easier to note that
$$
S=\sum_{n=1}^{2N} (-1)^{n-1}\frac1{n} = \sum_{n=1}^{N} \left(\frac1{2n-1} - \frac1{2n} \right)
$$
for then:
$$
S+\sum_{n=1}^{N} \frac1{n} = \sum_{n=1}^{2N} \frac1{n} = \sum_{n=1}^{N} \left( \frac1{n}+\frac1{N+n} \right)
$$
A: For LHS, use the definition of the skew harmonic number $\overline{H}_k=\sum_{n=1}^{k}\frac{(-1)^{n-1}}{n}$
$$\sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\overline{H}_{2N}=H_{2N}-H_N$$
For the RHS,
$$\sum_{n=1}^N\frac{1}{N+n}=\sum_{n=1+N}^{2N}\frac{1}{n}=\sum_{n=1}^{2N}\frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n}=H_{2N}-H_N$$
A: 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
$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}$$
