Find the sum of the series $\sum\frac{1}{(n+1)(n+2)(n+3)...(n+k)}$

Question

This question was in my maths paper . I tried to prove that the series converges but failed to do so. Any hint are welcomed .

Answer 1

$$\sum_{k = 1}^\infty\frac{1}{(n+1)(n+2)(n+3)…(n+k)} = n!\sum_{k = 1}^\infty\frac{1}{(n+k)!}$$ Which clearly converges super-exponentially. We can express this in terms of generalized Gamma function is desired.

Answer 2

Hint: $$(n+1)\cdot(n+2)\cdots(n+k)=\frac{(n+k)!}{n!}$$ from here you only need to get to $\displaystyle \sum \frac1{k!}=e$, which is easy by adding some missing terms.

Answer 3

Just focusing on convergence, since $n+k\geq n+2, \forall k\geq2,\forall n\geq0$ we have: $$0<\frac{1}{(n+1)(n+2)(n+3)...(n+k)}\leq\frac{1}{(n+1)(n+2)^{k-1}}$$ thus $$\sum\limits_{k=1}\frac{1}{(n+1)(n+2)(n+3)...(n+k)}\leq \frac{1}{n+1}+\sum\limits_{k=2}\frac{1}{(n+1)(n+2)^{k-1}}=\\ \frac{1}{n+1}\left(1+\sum\limits_{\color{red}{k=2}}\frac{1}{(n+2)^{k-1}}\right)= \frac{1}{n+1}\left(1+\sum\limits_{\color{red}{k=1}}\frac{1}{(n+2)^{k}}\right)=\\ \frac{1}{n+1}\left(\sum\limits_{\color{red}{k=0}}\frac{1}{(n+2)^{k}}\right)=...$$ the latter is an infinite geometric progression, or: $$...=\frac{1}{n+1}\cdot \frac{1}{1-\frac{1}{n+2}}=\frac{n+2}{(n+1)^2}$$