<s>[INST] Prove by induction that: $\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i}$ Basis step:
For $n=1$, equation holds.
Inductive step: Now,
 $$\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i} \tag{i. h.}$$
Now we want to show that
 $$\sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^{n+1}_{i=1}\frac{1}{n+1+i}.$$
So, 
$$ \sum^{2n+2}_{i=1} \frac{(-1)^{i-1}}{i} = \sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i} + \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2} = \text{(by inductive hypothesis)}= \sum^n_{i=1}\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2} $$
This is where I'm stuck.
 [/INST] Change the indices in the sum: $$\sum_{i=1}^n\frac{1}{n+i}+\frac{1}{2n+1}-\frac{1}{2n+2}=\sum_{i=0}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\
=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{n+1}+\frac{1}{2n+1}-\frac{1}{2n+2}\\=\sum_{i=1}^{n-1}\frac{1}{n+i+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\sum_{i=1}^{n+1}\frac{1}{n+i+1}.$$
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