Let I_n = int_0^1 e^{-y} y^n , dy,where n is | vedclass

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(C) We are given $I_n = \int_0^1 e^{-y} y^n \, dy$.Using the definition of the sum,we have $\sum_{n=1}^{\infty} \frac{I_n}{n!} = \sum_{n=1}^{\infty} \

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$\frac{1}{e}$

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(C) We are given $I_n = \int_0^1 e^{-y} y^n \, dy$.Using the definition of the sum,we have $\sum_{n=1}^{\infty} \frac{I_n}{n!} = \sum_{n=1}^{\infty} \frac{1}{n!} \int_0^1 e^{-y} y^n \, dy$.Since the integral is convergent,we can interchange the summation and the integral:$\sum_{n=1}^{\infty} \frac{I_n}{n!} = \int_0^1 e^{-y} \left( \sum_{n=1}^{\infty} \frac{y^n}{n!} \right) \, dy$.We know that the Taylor series for $e^y$ is $e^y = \sum_{n=0}^{\infty} \frac{y^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{y^n}{n!}$.Therefore,$\sum_{n=1}^{\infty} \frac{y^n}{n!} = e^y - 1$.Substituting this into the integral:$\sum_{n=1}^{\infty} \frac{I_n}{n!} = \int_0^1 e^{-y} (e^y - 1) \, dy = \int_0^1 (1 - e^{-y}) \, dy$.Evaluating the integral:$= [y - (-e^{-y})]_0^1 = [y + e^{-y}]_0^1$.$= (1 + e^{-1}) - (0 + e^0) = 1 + \frac{1}{e} - 1 = \frac{1}{e}$.

Let $I_n = \int_0^1 e^{-y} y^n \, dy$,where $n$ is a non-negative integer. Then,$\sum_{n=1}^{\infty} \frac{I_n}{n!}$ is

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