Sequence And Series Question 254
Question: The sum of the infinite series $ \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+…… $ is
[AMU 1999]
Options:
A) $ e-2 $
B) $ \frac{2}{3}e-1 $
C) 1
D) 3/2
Show Answer
Answer:
Correct Answer: C
Solution:
$ T_{n}=\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!} $ = $ \frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{(n)!}-\frac{1}{(n+1)!} $
$ \therefore T_1+T_2+T_3+….. $ = $ ( \frac{1}{1!}-\frac{1}{2!} )+( \frac{1}{2!}-\frac{1}{3!} )+( \frac{1}{3!}-\frac{1}{4!} )+…… $ = $ ( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+…… ) $ $ -( \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+…… ) $ = $ ( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+….-1 ) $ $ -( 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+….-1-1 ) $ = $ (e-1)-(e-2) $ = 1.
BETA
