# real analysis – How to compute the limit of the infinite series \$sum_{n=1}^{infty} frac{1}{n(n+1)ldots(n+m)}\$, \$m in mathbf{N}\$

For having some fun with infinite series, I am having a go at some of the exercises in `Problems in Mathematical Analysis` by Kaczor and Nowak. I would like to ask, if my approach is in the correct direction or if there’s another clever way of algebraically manipulating this. Please don’t post the solution – but rather a hint/tip.

Problem 3.1.4(a)

begin{align*} sum_{n=1}^{infty}frac{1}{n(n+1)(n+2)ldots(n+m)}, quad m in mathbf{N} end{align*}

Solution.

My approach would be to somehow, express this infinite sum as a telescopic series. Let’s start with a simpler case $$n=3$$. We would like to determine coefficients $$A,B,C,D$$ such that:

$$frac{1}{n(n+1)(n+2)(n+3)}= frac{A}{n} + frac{B}{n+1} + frac{C}{n+2} + frac{D}{n+3}$$

So,

begin{align*} 1 = A(n+1)(n+2)(n+3) + Bn(n+2)(n+3) + Cn(n+1)(n+3) + Dn(n+1)(n+2) end{align*}

Substituting the values $$n=0,-1,-2,-3$$, we find that, $$A=frac{1}{6},B=-frac{1}{2}, C=frac{1}{2}, D=-frac{1}{6}$$. Thus,

$$frac{1}{n(n+1)(n+2)(n+3)}= frac{1}{3!}left(frac{1}{n} – frac{3}{n+1} + frac{3}{n+2} – frac{1}{n+3}right)$$

Similarly,

$$frac{1}{n(n+1)(n+2)(n+3)(n+4)}= frac{1}{4!}left(frac{1}{n} – {4 choose 1} frac{1}{n+1} + {4 choose 2}frac{1}{n+2} – {4 choose 3}frac{1}{n+3} + frac{1}{n+4}right)$$

In general,

$$frac{1}{n(n+1)(n+2)(n+3)ldots(n+m)}= frac{1}{m!}left(frac{1}{n} – {m choose 1} frac{1}{n+1} + {m choose 2}frac{1}{n+2} – {m choose 3}frac{1}{n+3} + ldots+ frac{1}{n+m}right)$$

(This should be proved rigorously using mathematical induction.)

But, if now look at the partial sum $$s_n$$ and write it for simplicity as –

begin{align*} s_n &= frac{1}{m!} left(1 – {m choose 1} frac{1}{2} + {m choose 2}frac{1}{3} – {m choose 3 } frac{1}{4} + ldots + (-1)^mfrac{1}{1+m}right)\ &+ frac{1}{m!} left(frac{1}{2} – {m choose 1} frac{1}{3} + {m choose 2}frac{1}{4} – {m choose 3 } frac{1}{5} + ldots + (-1)^mfrac{1}{2+m}right)\ &+ vdots \ &+ frac{1}{m!} left(frac{1}{n} – {m choose 1} frac{1}{n+1} + {m choose 2}frac{1}{n+2} – {m choose 3 } frac{1}{n+3} + ldots + (-1)^mfrac{1}{n+m}right)\ end{align*}

This doesn’t really yield anything familiar, even if collect the terms for each fraction $$frac{1}{2},frac{1}{3},ldots$$.