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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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