Q:

Prove without induction : $\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^n \frac{1}{k+n}$

Prove without induction that : $$ 
\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^n \frac{1}{k+n} 
$$
Please if you have any elementary tricks just post hints.

A:

Rewriting $\displaystyle H(2n)-H(n)=\sum_{k=1}^n\frac1{k+n}$, the first part of this answer becomes:
$$
\begin{align}
\sum_{k=1}^n\frac1{k+n}
&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\
&=\sum_{k=1}^n\frac1{2k-1}+\frac1{2k}-\frac1k\\
&=\sum_{k=1}^n\frac1{2k-1}-\frac1{2k}\\
&=\sum_{k=1}^{2n}(-1)^{k-1}\frac1k
\end{align}
$$
which is the desired equation.

