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Question: Answered & Verified by Expert
Let $I_{n}=\int_{0}^{1} e^{-y} y^{n} d y$, where $n$ is a non-negative integer. Then $\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{I}_{\mathrm{n}}}{\mathrm{n} !}$ is
MathematicsDefinite IntegrationJEE Main
Options:
  • A 1
  • B $1-\frac{1}{\mathrm{e}}$
  • C $\frac{1}{\mathrm{e}}$
  • D $1+\frac{1}{\mathrm{e}}$
Solution:
2928 Upvotes Verified Answer
The correct answer is: $\frac{1}{\mathrm{e}}$
$I_{n}=\int_{0}^{1} e^{-y} y^{n} d y$
$I_{n}=-\frac{1}{e}+n I_{n-1}$ (by reduction formula)
$I_{n}-n I_{n-1}=-\frac{1}{e}$
$\begin{array}{l}
\frac{I_{n}}{n !}-\frac{I_{n-1}}{n-1 !}=-\frac{I}{e(n !)} \\
n=1 \quad \frac{I_{1}}{1 !}-\frac{I_{0}}{0 !}=-\frac{1}{e} \\
n=2 \quad \frac{I_{2}}{2 !}-\frac{I_{1}}{1 !}=-\frac{1}{e(2 !)} \\
n=3 \quad \frac{I_{3}}{3 !}-\frac{I_{2}}{2 !}=-\frac{1}{e(3 !)} \\
n=4 \quad \frac{I_{4}}{4 !}-\frac{I_{3}}{3 !}=-\frac{1}{e(4 !)} \\
\text { eq. }(1)+e q \cdot(2) \times 2+\operatorname{eq} \cdot(3) \times 3+(\text { eq. } 4) \times 4+\ldots \ldots \\
-\left\{\frac{I_{1}}{1 !}+\frac{I_{2}}{2 !}+\frac{I_{3}}{3 !}+\frac{I_{4}}{4 !}\right\}-\mathrm{I}_{0} \\
=-\frac{1}{e}\left\{1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots\right\}
\end{array}$
Let $\mathrm{S}=\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{I}_{\mathrm{n}}}{\mathrm{n} !}$
$\begin{array}{l}
-\mathrm{S}-\mathrm{I}_{0}=-\frac{1}{\mathrm{e}} \times \mathrm{e} \\
\mathrm{S}=1-\mathrm{I}_{0} \\
\mathrm{~S}=1-\int_{0}^{1} \mathrm{e}^{-\mathrm{y}} \mathrm{dy} \\
\mathrm{S}=1+\left[\mathrm{e}^{-\mathrm{y}}\right]_{0}^{1} \\
\mathrm{~S}=1+\left[\frac{1}{\mathrm{e}}-1\right] \\
\mathrm{S}=\frac{1}{\mathrm{e}}
\end{array}$

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