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

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(A) The $n$th term of the series is given by $T_n = \frac{1+2+3+\ldots+n}{(n+1) !}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(

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$\frac{e}{2}$

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(A) The $n$th term of the series is given by $T_n = \frac{1+2+3+\ldots+n}{(n+1) !}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(n+1)!} = \frac{n}{2(n)!} = \frac{1}{2(n-1)!}$.For $n=1, 2, 3, \ldots$,the terms are $T_1 = \frac{1}{2(0!)}, T_2 = \frac{1}{2(1!)}, T_3 = \frac{1}{2(2!)}, \ldots$.The sum $S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \left[ \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots \right]$.Since $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots$,we have $S = \frac{1}{2} e$.

The sum of the series $\frac{1}{2 !} + \frac{1+2}{3 !} + \frac{1+2+3}{4 !} + \ldots$ is equal to :

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