frac{1}{2!} + frac{1+2}{3!} + frac{1+2+3}{4!} | vedclass

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

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$\frac{e}{2}$

Vedclass · Answer

(A) The $n$-th term of the series is $T_n = \frac{1+2+3+\ldots+n}{(n+1)!}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(n+1)!} = \frac{n(n+1)}{2(n+1)n(n-1)!} = \frac{1}{2(n-1)!}$.Summing from $n=1$ to $\infty$:$S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{2(n-1)!} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$.Let $k = n-1$,then as $n=1, k=0$ and as $n 	o \infty, k 	o \infty$.$S = \frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{2} \left( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots ight) = \frac{1}{2} e$.Thus,the sum is $\frac{e}{2}$.

$\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots$ is equal to :

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