For each positive integer n, define f_ n (x)= Min ( x^ n n! , (1-x)^ n n! ) for 0 x 1 …
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
For each positive integer n, define
\[f_{n}(x)=\operatorname{Min}\left(\frac{x^{n}}{n!}, \frac{(1-x)^{n}}{n!}\right)\] for \(0 \leq x \leq 1\). Let
\(I_{n}=\int_{0}^{1} f_{n}(x) d x\) for \(n \geq 1\). Then the value of \(\sum_{n=1}^{\infty} I_{n}\) is
\[f_{n}(x)=\operatorname{Min}\left(\frac{x^{n}}{n!}, \frac{(1-x)^{n}}{n!}\right)\] for \(0 \leq x \leq 1\). Let
\(I_{n}=\int_{0}^{1} f_{n}(x) d x\) for \(n \geq 1\). Then the value of \(\sum_{n=1}^{\infty} I_{n}\) is
- \(2 \sqrt{e}-3\)
- \(2 \sqrt{e}-2\)
- \(2 \sqrt{e}-1\)
- \(2 \sqrt{e}\)
Answer
(A) 2 e -3
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