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College Mathematics

Sum Of Series: IIT JAM 2018 Problem 13

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

[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
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 ????

Connected Program at Cheenta

College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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Categories
College Mathematics

Sequences & Subsequences : IIT 2018 Problem 10

What are we learning?

Sequences & Subsequences are the key features in the filed of real analysis. We will see how to imply these concepts in our problem

Try to answer this question

[h5p id="10"]

Understand the problem

Let \(s_n\) = 1+\(\frac{1}{1!}\)+\(\frac{1}{2!}\)+……..+\(\frac{1}{n!}\) for n \(\in\) \(\mathbb{N}\) Then which of the following is TRUE for the sequence {\(s_{n}\}^\infty_{n=1}\):   (a) {\(s_{n}\}^\infty_{n=1}\) converges in \(\mathbb{Q}\).   (b) {\(s_{n}\}^\infty_{n=1}\) is a Cauchy sequence but does not converges to \(\mathbb{Q}\).   (c) The subsequence  {\(s_{k^n}\}^\infty_{n=1}\) is convergent in \(\mathbb{R}\) when k is a even natural number.   (d) {\(s_{n}\}^\infty_{n=1}\) is not a Cauchy sequence.
Source of the problem
IIT Jam 2018
Key competency
Gradient
Difficulty Level
Easy

Start with hints

Do you really need a hint? Try it first!

I am going to give you 3 clues in the beginning you try to work out using them. Then I will elaborate this clues in the following hints  (I) Every convergent sequence is a Cauchy sequence  (II)Every subsequence of a convergent sequence is convergent  (III)Consider then term 1+\(\frac{1}{1!}\)+\(\frac{1}{2!}\)+……..+\(\frac{1}{n!}\) Does this remind you any well known series?                                

I wil start with (III) consider \(e^x\)=1+\(\frac{x}{1!}\)+\(\frac{x^2}{2!}\)+……..+\(\frac{x^n}{n!}\) Isn’t the seris that we have to , is the value at x=1. Hence the given series\(\rightarrow\) e \(\in\) \(\mathbb{R}\) \ \(\mathbb{Q}\)

So option (a) is incorrect.

Every subsequence of a convergent sequence is convergent so {\(s_{k^n}\}^\infty_{n=1}\) is convergent not only for even k, but for any \(k \in \Bbb N\). So option (c) is incorrect.

Every convergent sequence is a Cauchy sequence so option (d) is incorrect and \(e \in\) \(\mathbb{R}\) so the given subsequence is convergent in \(\mathbb{R}\). So only option (b) is correct.

Look at the knowledge graph…

Play with graph

Fun fact: Do you know that this man has a sequence named after him?

Augustin-Louis Cauchy

Connected Program at Cheenta

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