The sum of the infinite series 1+frac{1}{2!}+ | vedclass

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Question

(C) Let the given series be $S = 1 + \frac{1}{2!} + \frac{1 \cdot 3}{4!} + \frac{1 \cdot 3 \cdot 5}{6!} + \dots$ The general term $T_n$ for $n \ge 1$

Vedclass · Accepted Answer

$\sqrt{e}$

Vedclass · Answer

(C) Let the given series be $S = 1 + \frac{1}{2!} + \frac{1 \cdot 3}{4!} + \frac{1 \cdot 3 \cdot 5}{6!} + \dots$ The general term $T_n$ for $n \ge 1$ is given by $T_n = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{(2n)!}$. We can rewrite the numerator as $\frac{(2n)!}{2^n n!}$. Thus, $T_n = \frac{(2n)!}{2^n n! (2n)!} = \frac{1}{2^n n!}$. The series is $S = \sum_{n=0}^{\infty} \frac{1}{2^n n!}$, where the $n=0$ term is $1$. This is the expansion of $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ with $x = \frac{1}{2}$. Therefore, $S = e^{1/2} = \sqrt{e}$.

The sum of the infinite series $1+\frac{1}{2!}+\frac{1 \cdot 3}{4!}+\frac{1 \cdot 3 \cdot 5}{6!}+\dots$ is

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