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

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

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