Is this sum integer? $b! \pi+1/(b+1)+1/(b+2)(b+1)+...$, where $b \neq 0$.

Question

Is $S$ an integer ?

$S= \: b! \:\pi+\frac{1}{b+1}+\frac{1}{(b+2)(b+1)} + \frac{1}{(b+3)(b+2)(b+1)} + ...$

$b \neq 0$.

Also, from here Is this sum rational or not? $1/(q+1)+1/(q+2)(q+1)...$ where $q$ is an integer

$\frac{1}{b+1}+\frac{1}{(b+2)(b+1)} + \frac{1}{(b+3)(b+2)(b+1)} + ...$ is irrational and between $(0,1)$.

Answer 1

Hint:

$$S_b=\frac{1}{b+1}+\frac{1}{(b+2)(b+1)} + \frac{1}{(b+3)(b+2)(b+1)} + ...+\: b! \:\pi=\dfrac{1}{b+1}(1+\frac{1}{b+2}+\frac{1}{(b+3)(b+2)} + \frac{1}{(b+4)(b+3)(b+2)} + ...+\: (b+1)! \:\pi)=\dfrac{S_{b+1}+1}{b+1}\to\\S_b=\dfrac{S_{b+1}+1}{b+1}$$can you finish now?

Answer 2

Turning a comment into an answer, note that if $0\lt b\in\mathbb{N}$, then

$$e=1+1+{1\over2}+{1\over6}+\cdots+{1\over b!}+{1\over(b+1)b!}+{1\over(b+2)(b+1)b!}+\cdots$$

and thus

$$S=b!(\pi+e)-B$$

where

$$B(b)=b!\left(1+1+{1\over2}+{1\over6}+\cdots+{1\over b!}\right)\in\mathbb{Z}$$

So the only way $S$ could be an integer is if $b!(\pi+e)$ were an integer. This would be the case for all large $b$ if $\pi+e$ were rational, and, of course, would never be the case if $\pi+e$ were irrational. The last I looked, $\pi+e$ is thought, but not known, to be irrational. So $S$ could possibly be an integer, but the smart money would say it isn't.