A) \[e\]
B) \[2\,e\]
C) e/2
D) None of these
Correct Answer: C
Solution :
\[S=\frac{1}{2\,!}+\frac{1+2}{3\,!}+\frac{1+2+3}{4\,!}+....\infty \] Here \[{{T}_{n}}=\frac{1+2+......+n}{(n+1)\ !}=\frac{\frac{n}{2}(n+1)}{(n+1)\ !}=\frac{1}{2(n-1)\ !}\] \[\Rightarrow S=\sum\limits_{n=1}^{\infty }{{{T}_{n}}=\frac{1}{2}}\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)\ !}=\frac{1}{2}e}\].
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