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 correctConnected 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
Related
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 correctConnected 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
Related
