MCQ
Let $I_n=\int_0^1 e^{-y} y^n d y$, where $n$ is a non-negative integer. Then, $\sum_{n=1}^{\infty} \frac{I_n}{n !}$ is
- A$ 1$
- B$1-\frac{1}{e}$
- ✓$\frac{1}{e}$
- D$1+\frac{1}{e}$
We have, $J_n=\int_0^1 e^{-y} y^n d y$
$I_n =\left[-y^n e^{-y}\right]_0^1+\int_0^1 n y^{n-1} e^{-y} d y$
$I_n =-\frac{1}{e}+n I_{n-1}$
$\Rightarrow \frac{I_n}{n !} =\frac{-1}{n ! e}+\frac{I_{n-1}}{(n-1) !}$
$\Rightarrow \frac{I_n}{n !}-\frac{I_{n-1}}{(n-1) !}=\frac{-1}{n ! e}$
$n =1 \frac{I_1}{1 !}-\frac{I_0}{0 !}=-\frac{1}{e}$
$n =2 \frac{I_2}{2 !}-\frac{I_1}{1 !}=-\frac{1}{2 ! e}$
$n =3 \frac{I_3}{3 !}-\frac{I_1}{2 !}=-\frac{1}{3 ! e}$
$n =n \frac{I_n}{n !}-\frac{I_n}{n-1}=\frac{1}{n ! e}$
$\text { Adding these form, we get }$
$\frac{I_n}{n !}-I_0 =-\frac{1}{e}\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots+\frac{1}{n !}\right)$
$\frac{I_n}{n !} =\frac{I_0}{n !}-\frac{1}{e}\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots+\frac{1}{n !}\right)$
$n ! =\frac{1}{e}+1-\frac{1}{e}(e)\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$
$=\frac{1}{e}+1-1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x)=\left\{\begin{array}{ll} \frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{array}\right.$
is continuous at $x=0,$ where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $X$.
Then :