For $f(m,n) = (1 - \frac{1}{m}) (1 - \frac{1}{m + 1}) \dots(1 - \frac{1}{n})$, find this sum

Question

Let $m,n$ be integers and $1 < m \leq n$ .

Define : $f(m,n) = \left(1 - \frac{1}{m}\right) \left(1 - \frac{1}{m + 1}\right)\left(1 - \frac{1}{m + 2}\right)... \left(1 - \frac{1}{n}\right) .$

If $S = f(2,2049) + f(3,2049) + f(4,2049) + ... + f(2049,2049)$ ; find $\frac{S}{64}$ .

What I Think :- I think this can be done by telescoping sum but I can't understand how . A part of this was posted here, from here I can say that $f(2,2049)$ $=$ $\frac{1}{2049}$ (and it's easy to find that), but what about the other terms? Can anyone help me?

Answer 1

$f(m,n)=\frac{(m-1)(m)....(n-1)}{(m)(m+1)...(n)}=\frac{(m-1)}{n}$

your sum is $$S = f(2,2049) + f(3,2049) + f(4,2049) + ... + f(2049,2049)$$ which can be rewritten as $$\frac{2-1}{2049}+\frac{3-1}{2049}...+\frac{2048}{2049}$$ $$\sum_{m=2}^{2049}\frac{m-1}{2049}$$ which can be evaluated easily