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(D) The $n$-th term of the series is given by $T_n = \frac{1 + x + x^2 + \dots + x^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{1

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$\frac{e^x - e}{x - 1}$

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(D) The $n$-th term of the series is given by $T_n = \frac{1 + x + x^2 + \dots + x^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{1 - x^n}{1 - x} \cdot \frac{1}{n!} = \frac{x^n - 1}{(x - 1)n!} = \frac{1}{x - 1} \left( \frac{x^n}{n!} - \frac{1}{n!} \right)$.Summing from $n = 1$ to $\infty$,we get $\sum_{n=1}^{\infty} T_n = \frac{1}{x - 1} \left( \sum_{n=1}^{\infty} \frac{x^n}{n!} - \sum_{n=1}^{\infty} \frac{1}{n!} \right)$.Since $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n!} = e - 1$,$\sum_{n=1}^{\infty} T_n = \frac{1}{x - 1} ((e^x - 1) - (e - 1)) = \frac{e^x - e}{x - 1}$.

$1 + \frac{1 + x}{2!} + \frac{1 + x + x^2}{3!} + \frac{1 + x + x^2 + x^3}{4!} + \dots \infty = $

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