Warm Yourself With An MCQ
Understand the problem
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 (\(a_n\))=3Source of the problem
IIT JAM 2018 Question 11
Topic
Infinum and Supremum
Difficulty Level
Easy
Suggested Book
Real Analysis By S K Mapa
Start with hints
Do you really need a hint? Try it first!
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?
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?
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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