Supremum and Infimum: IIT JAM 2018 Problem 11
Understand the problem
$ a_n=\begin{cases}
2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\
1+ \frac{1}{2^n}, & \text{if n is even}
\end{cases}$
Which of the following is true?
(a) sup {\(a_n|n \in \mathbb{N}\)}=3 and inf {\(a_n|n \in \mathbb{N}\)}=1
(b) lim inf (\(a_n\))=lim sup (\(a_n\))=\(\frac{3}{2}\)
(c) sup {\(a_n|n \in \mathbb{N}\)}=2 and inf {\(a_n|n \in \mathbb{N}\)}=1
(d) lim inf (\(a_n\))= 1 lim sup (\(a_n\))=3
Start with hints
Hint 1:
$a_n=\begin{cases}
2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\
1+ \frac{1}{2^n}, & \text{if n is even}
\end{cases}$
Now the limit points of this set are those points which the set does not attain.So, they might be the sup and inf which are not attained by this set.
Basically sup(\(a_n\))= max{ limit points, \(a_n\) | n \(\in\) \(\mathbb{N}\)}
Limit points are \(2,1\) and \(a_1= 2+1=3, a_3= 2- \frac{1}{3} ; a_5= 2+\frac{1}{5} \)
\(a_0= 1+1=2 , a_2= 1+ \frac{1}{4} , a_3= 1+\frac{1}{8} \)
Now you can calculate the supremum?
Hint 2:
From the observation of Hint 2 we have
sup \(a_n\)= max \(\{2,1,3,2\}=3 \)
Similarly, inf \(a_n\)= min\(\{\) limit points, \(a_n | n \in \mathbb{N}\}\)
Can you calculate that by yourself?
Hint 3:
inf \(a_n\)= min {2,1,2 -\(\frac{1}{3}\)}=1
So, option A is correct.
Now there is another question regarding lim sup and lim inf. We can observe that we have mainly \(3\) subsequences , corresponding to
\( n\) is even; \(n=2k\)
\(n\)= \(4k+1\)
\(n=4k+3\)
Can you calculate the corresponding subsequences and their limits?
Hint 4:
For \(n=2k\) we have
\(a_{2k}=1+ \frac{1}{2^{ek}} \longrightarrow 1 \) ask
For \(a_{4k+1}= 2+ \frac{1}{4k+1} \longrightarrow 2\) ask
\(a_{4k+3}= 2-\frac{1}{4k+3} \longrightarrow 2\) ask
So, lim sup \(a_n\)=max\(\{1,2\}=2\)
Lim inf \(a_n\)=min\(\{1,2\}=1\)
Therefore, Option C is also correct
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Similar Problems
Sequences & Subsequences : IIT 2018 Problem 10
What are we learning?
Sequences, Subsequences are the key features in the field of real analysis. We will see how to imply these concepts in our problem
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. Difficulty Level Easy Suggested Book
Start with hints
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.
Connected Program at Cheenta
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.