This is a question I stumbled upon while learning sequence and series
Is $$\sum_{r=1}^n\frac1{r(r+1)(r+2)(r+3)...(r+m)}$$ the same as $$\frac1{m}\left(\frac1{1\times 2\times \cdots \times (m-1)}-\frac1{(n+1)(n+2)\cdots(n+m)}\right)?$$
A summary of what I tried so far is that I tried writing the denominators of the LHS into series and multiply to form a bigger series but the LHS after simplification did not convert to the RHS.I also tried to write the LHS as an AGP but it didn't work out. So I would request you to please guide me with a hint, not a solution.
The answer is that both expressions are identical in terms of magnitude
Please note this is not my homework exercise, I am just learning algebra for my competitive examination preparation.
EDIT- Telescopicing sum will solve the question easily.
Hint: Telescoping sum and $$\frac{m}{r(r+1)(r+2)\cdots(r+m)}=\frac{1}{r(r+1)\cdots(r+m-1)}-\frac{1}{(r+1)(r+2)\cdots(r+m)}.$$