The sum of the series frac{1}{1 times 2} + fr | vedclass

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(C) The $n^{th}$ term of the given series is $T_n = \frac{1 	imes 3 	imes 5 	imes \dots 	imes (2n - 1)}{(2n)!}$.Multiplying the numerator and deno

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$e^{1/2} - 1$

Vedclass · Answer

(C) The $n^{th}$ term of the given series is $T_n = \frac{1 	imes 3 	imes 5 	imes \dots 	imes (2n - 1)}{(2n)!}$.Multiplying the numerator and denominator by $2 	imes 4 	imes 6 	imes \dots 	imes (2n)$,we get:$T_n = \frac{(2n)!}{(2n)! 	imes 2^n 	imes n!} = \frac{1}{2^n 	imes n!} = \frac{(1/2)^n}{n!}$.The sum of the series is $S = \sum_{n=1}^{\infty} \frac{(1/2)^n}{n!}$.Using the expansion $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}$,we have $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$.Substituting $x = 1/2$,we get $S = e^{1/2} - 1$.

The sum of the series $\frac{1}{1 \times 2} + \frac{1 \times 3}{1 \times 2 \times 3 \times 4} + \frac{1 \times 3 \times 5}{1 \times 2 \times 3 \times 4 \times 5 \times 6} + \dots \infty$ is

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