Sum of an infinite series I have no idea how to compute the sum of the following series:
$ \sum_{k=b+1}^{\infty}\left(\frac{b!}{k!}+\frac{6b!}{k^2}\right) $. Could you help me? I tried even integrals, but I have still nothing.
 A: First part:
$$b!\sum_{k=b+1}^{\infty} \frac 1 {k!}=b!(\sum_{k=0}^{\infty} \frac 1 {k!}-\sum_{k=0}^{b} \frac 1 {k!})=b!(e-e\frac{\Gamma(b+1,1)}{\Gamma(b+1)})
$$
where $ \Gamma(b+1,1)$ is the incomplete Gamma function. 
Second part:
$$6(b!) \sum_{k=b+1}^{\infty}\frac 1 {k^2}=6(b!)(\sum_{k=1}^{\infty}\frac 1 {k^2}-\sum_{k=1}^{b}\frac 1 {k^2})=6(b!)(\frac{\pi^2}{6}-H^{(2)}_b)$$
where $H^{(2)}_b$ is the generalized harmonic number.
A: For a general $b
 $ we have $$\sum_{k=b+1}^{\infty}\left(\frac{b!}{k!}+\frac{6b!}{k^{2}}\right)=b!\left(\sum_{k=0}^{\infty}\frac{1}{k!}-\sum_{k=0}^{b}\frac{1}{k!}\right)+6b!\left(\sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{k=1}^{b}\frac{1}{k^{2}}\right)=
 $$ $$=b!e-e\Gamma\left(b+1,1\right)+6b!\left(\frac{\pi^{2}}{6}-H_{b}^{\left(2\right)}\right)
 $$ where $\Gamma\left(a,x\right)
 $ is the incomplete Gamma function and $H_{n}^{\left(r\right)}
 $ is the generalized harmonic number.
A: You wanted to show that this is not an integer. I will show you that this is not a rational number. Suppose it is rational. Then
$$\sum_{k\ge b+1} \left( \frac{b!}{k!} + 6\frac{b!}{k^2} \right) + \sum_{1\le k\le b} \left( \frac{b!}{k!} + 6\frac{b!}{k^2} \right) = b!(e-1 + \pi^2)$$
would be rational since the second summand is obviously rational.
Multiply everything with $\frac{1}{b!}$ and add $1$.
You just showed that $e+\pi^2$ is rational: a contradiction.
