Question

Let $1_n={\int }_0^1{e}^{-y}{y}^{n}dy$, where n is a non-negative integer. Then $\sum _{n=1}^{\infty }\frac{I_n}{n!}$ is?

• A. $1$
• B. $1-\frac{1}{e}$
• C. $\frac{1}{e}$
• D. $1+\frac{1}{e}$

Answer 1

Answer: (C)

$\frac{1}{e}$

$I_n={\int }_0^1{e}^{-y}{y}^{n}dy$

$⟹I_n=y^n{\int }_0^1{e}^{-y}dy-{\int }_0^1\left(n{y}^{n-1}\right)\{\int e^{-y}dy\}dy$

$⟹I_n=-e^{-y}{y}^n{|}_0^1+n{\int }_0^1\left(e^{-y}\right)y^{n-1}dy$

$⟹I_n=-e^{-1}\times 1^n+0+n{I}_{n-1}$

$⟹I_n=-e^{-1}+n{I}_{n-1}$

$⟹I_n=-e^{-1}+\frac{n!}{(n-1)!}{I}_{n-1}$

$⟹I_n=-e^{-1}+\frac{n!}{(n-1)!}(-e^{-1})+\frac{n!}{(n-2)!}(-e^{-1})\cdots +\frac{n!}{1!}{1}_1\dots \left(1\right)$

Also,

$I_1={\int }_0^1{e}^{-y}ydy$

$⟹I_1=-e^{-y}y{|}_0^1-\int -e^{-y}dy$

$⟹I_1=1-\frac{2}{e}$

Now from equation (1), we have

$⟹I_n=-e^{-1}\{1+\frac{n!}{(n-1)!}+\frac{n!}{(n-2)!}\cdots +\frac{n!}{2!}\}+\frac{n!}{1!}\{1-\frac{2}{e}\}$

$⟹\frac{I_n}{n!}=-e^{-1}\{\frac{1}{n!}+\frac{1}{(n-1)!}+\frac{1}{(n-2)!}\cdots +\frac{1}{2!}+1\}+\frac{1}{e}+\{1-\frac{2}{e}\}$

Also,

$n!{\sum }_{k=1}^n\frac{1}{k!}=\lfloor n!e\rfloor -1\dots \left(2\right)$

$\therefore {\sum }_{n=1}^{\infty }\frac{I_n}{n!}={\sum }_{n=1}^{\infty }(-e^{-1})\{\frac{1}{n!}+\frac{1}{(n-1)!}+\frac{1}{(n-2)!}\cdots +\frac{1}{2!}+1\}+\frac{1}{e}+\{1-\frac{2}{e}\}$

$\therefore {\sum }_{n=1}^{\infty }\frac{I_n}{n!}=-\frac{1}{e}{\sum }_{n=1}^{\infty }\{\frac{1}{n!}+\frac{1}{(n-1)!}+\frac{1}{(n-2)!}\cdots +\frac{1}{2!}+1\}+e$

$\therefore {\sum }_{n=1}^{\infty }\frac{I_n}{n!}=\frac{1}{e}{\sum }_{n=1}^{\infty }-\frac{\lfloor n!e\rfloor -1}{n!}+e$ (from (2))

$\therefore {\sum }_{n=1}^{\infty }\frac{I_n}{n!}=\frac{1}{e}$