Summation of reciprocals

You can do it via induction, for $n=1$ it is true. Now suppose it holds for $n$, then use that

$$\sum_{i=1}^{n+1}\frac1{n+1+i}=\sum_{i=1}^{n}\frac1{n+i}+\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}$$

and

$$\sum_{i=1}^{2(n+1)}\frac{(-1)^{i+1}}{i}=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}-\frac{1}{2n+2}$$

by induction when you take the difference the sums drop out and you are left with

$$\left(\frac1{2n+1}+\frac{1}{2n+2}-\frac1{n+1}\right)-\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\frac{2}{2n+2}-\frac{1}{n+1}=0$$

For an an alternative direct proof, start with the right hand side sum and split then recombine the odd and even terms, respectively.

$$ \begin{align} \sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i} & = \sum_{i=1}^{n}\frac{1}{2 i - 1} - \sum_{i=1}^{n}\frac{1}{2 i} \\ & = \sum_{i=1}^{n}\frac{1}{2 i - 1} - \frac{1}{2} \sum_{i=1}^{n}\frac{1}{i} \\ & = \sum_{i=1}^{n}\frac{1}{2 i - 1} + \frac{1}{2} \sum_{i=1}^{n}\frac{1}{i} - \sum_{i=1}^{n}\frac{1}{i} \\ & = \sum_{i=1}^{2 n}\frac{1}{i} - \sum_{i=1}^{n}\frac{1}{i} = \\ & = \sum_{i=n+1}^{2 n}\frac{1}{i} = \sum_{i=1}^{n}\frac{1}{n + i} \end{align} $$