<s>[INST] How to prove using induction that $ \sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n} = \frac{2n}{n+1}$ Using Mathematical Induction I need to prove that
$ \sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n}  =  \frac{2n}{n+1}$
As A first step I verified it for numbers 1 and 2 which worked
Secondly I simplified the LHS as:
$ \frac{1}{\sum n}  = \frac{1}{\frac {n(n+1)}{2}}  =  \frac{2}{n(n+1)}$
which converts my problem into
$ \frac{2}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{2}{3 \cdot 4} ......... + \frac{2}{n \cdot (n+1)} = \frac{2n}{n+1} $
After which I cancel out 2 as the common factor from both sides. But at this stage I am stuck and am unable to go forward. What should I do next ?
 [/INST] Straightforward. After simplifying your expression, proceed with your induction.
$$\sum_{i=1}^{n+1} \dfrac{2}{i(i+1)} = \sum_{i=1}^{n} \dfrac{2}{i(i+1)} + \frac{2}{(n+1)(n+2)} = \frac{2n}{n+1} + \frac{2}{(n+1)(n+2)} = \frac{2n^2+4n+2}{(n+1)(n+2)} = \frac{2(n+1)}{n+2}  $$
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