TITLE: Summation of reciprocals
QUESTION [9 upvotes]: $$\sum_{i=1}^n \frac1{n+i} = \sum_{i=1}^{2n}\frac{(-1)^{i+1}}i,\quad \text{ for }n\ge 1.$$
I really have no clue how on to solve this one. It's the last one on my assignment. Would really enjoy some help with this one.

REPLY [9 votes]: For an an alternative direct proof, start with the right hand side sum and split then recombine the odd and even terms, respectively.
$$
\begin{align}
\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i} & = \sum_{i=1}^{n}\frac{1}{2 i - 1} - \sum_{i=1}^{n}\frac{1}{2 i} \\
 & = \sum_{i=1}^{n}\frac{1}{2 i - 1} - \frac{1}{2} \sum_{i=1}^{n}\frac{1}{i} \\
 & = \sum_{i=1}^{n}\frac{1}{2 i - 1} + \frac{1}{2} \sum_{i=1}^{n}\frac{1}{i} -  \sum_{i=1}^{n}\frac{1}{i} \\
 & = \sum_{i=1}^{2 n}\frac{1}{i} -  \sum_{i=1}^{n}\frac{1}{i} = \\
 & = \sum_{i=n+1}^{2 n}\frac{1}{i} = \sum_{i=1}^{n}\frac{1}{n + i} 
\end{align}
$$