<s>[INST] Prove that $\sum_{k=1}^{m}\frac{1}{k(k+1)}=1-\frac{1}{m+1}$. I know $\sum_{k=1}^{m}\frac{1}{k(k+1)}=1-\frac{1}{m+1}= H_mH_{m+1}$,   for $H_m$ the harmonic sum.
I tried many ways to prove it like this 
$$\sum_{k=1}^{m}\left(\frac{1}{k(k+1)} + \frac{1}{m+1}\right)=1  $$  
$$ \Rightarrow \sum_{k=1}^{m-1} \left(\frac{1}{k(k+1)}+\frac{1}{m+1} + \frac{1}{m(m+1)}\right) = \sum_{k=1}^{m-1} \left(\frac{1}{k(k+1)} + \frac{m^2 + 1}{m(m+1)}\right) = 1.$$
My questions are:


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*Could anyone give me any hints to calculate it, but not the solution, please? 
Thanks! $\Rightarrow$ it's OK I solved the exercise :) !! 

*Is the calculation, I wrote above here, correct? Or not?
 [/INST] Induction would probably work out here too. For $m=1$ we have $$\frac1{1\cdot 2} = 1-\frac12.$$ Then for $m=2$ we have $$\frac{1}{1\cdot2} + \frac{1}{2\cdot 3} = \left(1 - \frac12\right) + \frac1{2\cdot3}=1-\frac{3-1}{2\cdot 3}=1-\frac{1}{3}.$$
Try to do this in general.
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