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)^m\frac{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)^m\frac{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)^m\frac{1}{n+m}\right]\\ \end{align*}
This doesn't really yield anything familiar, even if collect the terms for each fraction $\frac{1}{2},\frac{1}{3},\ldots$.
