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 ????

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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

Tom M. Apostol

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

College Mathematics Program

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