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# Sum Of Series: IIT JAM 2018 Problem 13

This problem appeared in IIT JAM 2018. This problem required some minor but very basic concepts of how to find the sum of a seris.

[h5p id="14"]

# Understand the problem

Let $a_n = n+\frac{1}{n} , n \in \mathbb{N}$. Then the sum of the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} is$ (A) $e^{-1}-1$ (B) $e^{-1}$ (C) $1-e^{-1}$ (D) $1+e^{-1}$
##### Source of the problem
IIT JAM 2018 Problem 13
Series
Easy
##### Suggested Book
Real Analysis By S.K Mapa

Do you really need a hint? Try it first!

Consider $a_n = n + \frac{1}{n} , n \in \mathbb{N}$ We have to use $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!}+.....$ Specifically $e^{1} = 1+ \frac{1}{1!} + \frac{1}{2!}+....$ And $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + .......$ Do you want to play with it $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1 + \frac{1}{n+1}}{n!}$ = $\sum_{n=1}^{\infty}[ (-1)^{n+1} \frac{1}{(n-1)!} + (-1)^{n+1} \frac{1}{n!} + (-1)^{n+1} \frac{1}{(n+1)!}]$   Now we will be breaking it term by term for the ease of calculation. Can you do it from here? $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!}$ = $[1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......]$ = $e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1]$ = $e^{-1} + 1$ So option (D) is our required answer.

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