JEE Advanced 2019 Paper 2 · Mathematics · Sequences and Series · medium

The sum of the infinite series Σn=1^∞(n^2+n+1)/((n+2)!) is ______.

Numerical answer

The sum of the infinite series

\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2+n+1}{(n+2)!}

is ______. (Give the answer correct to two decimal places.)

Show answer and solution

Answer: 0.94

Write

n^2+n+1=(n+2)(n+1)-2(n+1)+1

. Dividing by

(n+2)!

\frac{(n+2)(n+1)}{(n+2)!}=\frac1{n!}

\frac{n+1}{(n+2)!}=\frac{1}{(n+1)!}-\frac1{(n+2)!}

and the last term is

\frac1{(n+2)!}

. Therefore the general term is

\frac1{n!}-\frac2{(n+1)!}+\frac3{(n+2)!}

. Using

\sum_{n\ge1}\frac1{n!}=e-1

\sum_{n\ge1}\frac1{(n+1)!}=e-2

and

\sum_{n\ge1}\frac1{(n+2)!}=e-\frac52

, the sum is

(e-1)-2(e-2)+3\left(e-\frac52\right)=2e-\frac92\approx0.9366

, which is

0.94

to two decimals. Summing the first forty terms numerically gives

0.93656

, confirming it.

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