Q:

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:

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?

A:

Hint: $\frac1{k(k+1)}=\frac1k-\frac1{k+1}$.

