<s>[INST] Proving by induction that $\sum\limits_{i=1}^n\frac{1}{n+i}=\sum\limits_{i=1}^n\left(\frac1{2i-1}-\frac1{2i}\right)$ I have a homework problem to prove the following via induction:
$$\sum_{i=1}^n \frac{1}{n+i} = \sum_{i=1}^n \left(\frac{1}{2i-1} - \frac{1}{2i}\right) $$
The base case is true.
So far I've done the following:
$$\sum_{i=1}^{k+1} \frac{1}{(k+1)+i}$$
$$a_{k+1} = \frac{1}{(k+1)+(k+1)} = \frac{1}{2k +2} $$
$$s_k + a_{k+1} = \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) $$
$$s_{k+1} = \sum_{i=1}^{k+1} (\frac{1}{2(k+1)-1} - \frac{1}{2(k+1)}) $$
$$ \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{i=1}^{k+1} (\frac{1}{2(k+1)-1} - \frac{1}{2(k+1)}) $$
$$ \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{i=1}^{k+1} (\frac{1}{2k-1} - \frac{1}{2k+2)}) $$
I'm having a hard time figuring out how to proceed from here.  Anyone have any hints or tips for search terms? Or corrections to what I've started above?
 [/INST] Write the left hand side as $f(n)$.
Note that $$f(n+1)-f(n)=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$
That's the key step - that $f(n+1)$ adds two terms to the right of the sum, but subtracts one term from the left.
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