Prove Catalan's Identity I need to prove the following Identity 
$$\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n-1}+\frac1{2n}=1-\frac12+\frac13-\cdots+\frac1{2n-1}-\frac1{2n}$$
Which in compact form is: $$\sum_{j=1}^n \frac1{n+j}=\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}$$
I can also express the previous equality as:
$$\sum_{j=1}^n \frac1{n+j}=\sum_{j=1}^{n} \frac1{2j-1}-\sum_{j=1}^{n} \frac1{2j}$$
However I can't seem to go any further. Any hint is appreciated!
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{j = 1}^{n}{1 \over n + j} & =
\sum_{j = 1 + n}^{2n}{1 \over j} =
\sum_{j = 1}^{2n}{1 \over j} - \sum_{j = 1}^{n}{1 \over j} =
\bracks{\sum_{j = 1}^{n}{1 \over 2j} + \sum_{j = 1}^{n}{1 \over 2j - 1}} - 2\sum_{j = 1}^{n}{1 \over 2j}
\\[5mm] & =
\sum_{j = 1}^{n}{1 \over 2j - 1} - \sum_{j = 1}^{n}{1 \over 2j} =
\bbx{\ds{\sum_{j = 1}^{2n}{\pars{-1}^{n + 1} \over j}}}
\end{align}
