JEE Advanced 2018 Paper 1 · Mathematics · Sequences and Series · medium

Evaluate Σn=1^∞(n^2/((n+1)!), giving the answer correct to two decimal places.

Numerical answer

Evaluate

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

, giving the answer correct to two decimal places.

Show answer and solution

Answer: 1.72

Write

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

. Then

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

. Also

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

, which telescopes:

\sum_{n\ge1}\frac n{(n+1)!}=\frac1{1!}=1

. The first part gives

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

. Therefore the sum equals

e-1\approx1.71828

, which is

1.72

to two decimal places. A direct numerical partial sum up to

n=40

agrees with this value.

Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake