Infinite summation of recursive integral

Let $$I_n=\int_{0}^{1}e^{-y}y^n\ dy$$, where $$n$$ is non-negative integer. Find $$\sum_{n=1}^{\infty}\frac{I_n}{n!}.$$

I first solved $$I_n$$ and obtained $$I_n=-\frac{1}{e}+nI_{n-1} \\ \hspace{35mm} =-\frac{1}{e}+n(-\frac{1}{e}+(n-1)I_{n-2}) \\ \hspace{72.7mm}=-\frac{1}{e}-n\frac{1}{e}-n(n-1)\frac{1}{e}-n(n-1)\frac{1}{e}+...+n!(I_1)$$

Since $$I_1=1-\frac{2}{e}\\ \therefore \frac{I_n}{n!}=-\frac{1}{e}(\frac{1}{n!}+\frac{1}{(n-1)!}+\frac{1}{(n-2)!}...+\frac{1}{2})+(1-\frac{2}{e})$$

I don't know what to do after this, because taking $$e=2+1/2!+1/3!+1/4!+...+1/n!$$ gives the value of $$\frac{I_n}{n!}$$ to be $$0$$.

How do I solve this?

• $dx$ should be $dy$ and why are you not integrating from $0$ to $1$?
– Gary
Dec 12, 2021 at 2:17
• @Gary Typo, my bad. Dec 12, 2021 at 2:20

How about $$\sum\limits_{n = 1}^\infty {\frac{{I_n }}{{n!}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{n!}}\int_0^1 {e^{ - y} y^n dy} } = \int_0^1 {e^{ - y} \sum\limits_{n = 1}^\infty {\frac{{y^n }}{{n!}}} dy} = \int_0^1 {e^{ - y} (e^y - 1)dy} = e^{ - 1} \,?$$