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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?

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

1 Answer 1

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

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    $\begingroup$ I didn't know it was possible to bring the summation inside the integral! $\endgroup$
    – Alpha Delta
    Dec 12, 2021 at 2:32
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    $\begingroup$ You can use the monotone convergence theorem to justify it. $\endgroup$
    – Gary
    Dec 12, 2021 at 2:45

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