Summation Of Binomial & Factorial Series

Question

Looking for an explicit formula for the following:

$$ S = \sum _{i=j}^n \frac{\binom{i}{j}}{(i+1)!} $$

any Ideas?

Answer 1

I can't give a solution to it but my hint would be that $i$ pick $j$ is equal to $\frac{i!}{(i-j)!j!}$ which combined with what you had gives $$s=\sum_{i=j}^n \frac{1}{i(i-j)!j!}$$ Just some to think about.

Answer 2

We have

$$\begin{align}S &= \sum _{i=j}^n \frac{\binom{i}{j}}{(i+1)!}\\ &=\sum_{i=j}^n \frac{1}{(i+1)(i-j)!j!}\\ &=\frac{1}{j!}\sum_{k=0}^{n-j}\frac{1}{(k+j+1)k!}\end{align}$$

This does not have a closed form but with $n\to\infty$ the sum is a $j^{th}$ anti derivative of the exponential...

Answer 3

$$\sum_{i=j}^{n}\frac{\binom{i}{j}}{(i+1)!}=\frac{1}{j!}\sum_{i=j}^{n}\frac{1}{(i+1)(i-j)!}=\frac{1}{j!}\sum_{h=0}^{n-j}\frac{1}{(h+j+1)h!}$$ is just a partial sum for a hypergeometric function, but a nice upper bound exists: $$\sum_{i=j}^{n}\frac{\binom{i}{j}}{(i+1)!}\leq \frac{1}{j!}\int_{0}^{1}x^j e^x\,dx=e\left(\frac{1}{j!}-\frac{1}{(j-1)!}+\ldots\pm\frac{1}{0!}\right)-(-1)^j. $$