Sequence And Series Question 513
Question: $ \frac{1}{2,!}+\frac{1+2}{3,!}+\frac{1+2+3}{4,!}+……\infty = $
[EAMCET 2003]
Options:
A) $ e $
B) $ 2,e $
C) e/2
D) None of these
Show Answer
Answer:
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} $ .
BETA
