How Cheenta works to ensure student success?
Explore the Back-Story

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

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.

Similar Problems

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

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.

Similar Problems

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
magic-wandrockethighlight