For each positive integer n,define f_n(x) = m | vedclass

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(A) Given $f_n(x) = \min\left(\frac{x^n}{n!}, \frac{(1-x)^n}{n!}\right)$ for $x \in [0, 1]$.Since $\frac{x^n}{n!} \leq \frac{(1-x)^n}{n!}$ for $0 \leq

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$2\sqrt{e} - 3$

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(A) Given $f_n(x) = \min\left(\frac{x^n}{n!}, \frac{(1-x)^n}{n!}\right)$ for $x \in [0, 1]$.Since $\frac{x^n}{n!} \leq \frac{(1-x)^n}{n!}$ for $0 \leq x \leq \frac{1}{2}$ and $\frac{(1-x)^n}{n!} \leq \frac{x^n}{n!}$ for $\frac{1}{2} \leq x \leq 1$,we have:$I_n = \int_{0}^{1/2} \frac{x^n}{n!} dx + \int_{1/2}^{1} \frac{(1-x)^n}{n!} dx$.By symmetry,$I_n = 2 \int_{0}^{1/2} \frac{x^n}{n!} dx = 2 \left[ \frac{x^{n+1}}{(n+1)n!} \right]_{0}^{1/2} = 2 \frac{(1/2)^{n+1}}{(n+1)!} = \frac{2}{(n+1)! 2^{n+1}} = \frac{1}{(n+1)! 2^n}$.Now,$\sum_{n=1}^{\infty} I_n = \sum_{n=1}^{\infty} \frac{1}{(n+1)! 2^n}$.Let $k = n+1$,then $\sum_{k=2}^{\infty} \frac{(1/2)^{k-1}}{k!} = 2 \sum_{k=2}^{\infty} \frac{(1/2)^k}{k!}$.Since $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$,we have $\sum_{k=2}^{\infty} \frac{(1/2)^k}{k!} = e^{1/2} - (1 + 1/2) = \sqrt{e} - \frac{3}{2}$.Thus,$\sum_{n=1}^{\infty} I_n = 2(\sqrt{e} - \frac{3}{2}) = 2\sqrt{e} - 3$.

For each positive integer $n$,define $f_n(x) = \min\left(\frac{x^n}{n!}, \frac{(1-x)^n}{n!}\right)$ for $0 \leq x \leq 1$. Let $I_n = \int_{0}^{1} f_n(x) dx$ for $n \geq 1$. Then,$\sum_{n=1}^{\infty} I_n$ is equal to

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