# Compute $\frac{1}{e}\sum\limits_{n=0}^{\infty}\frac{n^{k}}{n!}$ for $k=0, 1, 2 …$

Using some matlab (I know it's cheating) I found that: $$k=0 => Result=1$$ $$k=1 => Result=1$$ $$k=2 => Result=2$$ $$k=3 => Result=5$$ $$k=4 => Result=15$$ $$k=5 => Result=52$$ $$k=6 => Result=203$$

This sequence is puzzling...

• Am I right that these are rounded results? – wythagoras May 4 '15 at 12:42
• Bell number – achille hui May 4 '15 at 12:50
• But you don't seem to be asking a question, Chris. – Gerry Myerson May 4 '15 at 13:01
• The results for k=0,1,2,3 are accurate. I computed the series by hand. The others are rounded, but they converged very quickly with accuracy of 10 zeros after the point. My question is what is the sequence $a_k=\frac{1}{e}\sum\limits_{n=0}^{\infty}\frac{n^{k}}{n!}$. achille hui recognized the numbers as Bell Numbers, but how do you prove it? – Chris May 4 '15 at 13:10
• It may interest you that this is the third time in two weeks this sum has appeared here i.e. at this MSE link. – Marko Riedel May 4 '15 at 19:44

$$\Large\sum_{n=1}^{\infty}\frac{\mathbb e^{nx}}{n!}=\mathbb e^{\mathbb e^x}-1$$
$$~$$
$$\Large\sum_{n=1}^\infty \frac{n^k}{n!}=\frac{d^k}{dx^k}(\mathbb e^{\mathbb e^x}-1)\huge]_{x=0}$$
• hmm...where's $j$? – Alex May 4 '15 at 13:31
• What is $\frac{e^{nx}}{!}$? – apnorton May 4 '15 at 13:38