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College Mathematics

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.

What about a small warm up MCQ!!!!!

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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
Topic
Series
Difficulty Level
Easy
Suggested Book
Real Analysis By S.K Mapa

Start with hints

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.

The will look more easy if we take a look into the knowledge graph

Let’s have a look into the graphs 

Do You Know ????

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