[College Calc 01-02. Limits of Sequences] #13

Calculus Level 4

Which of the followings are true for any bounded sequences { a n } , { b n } \{a_n\},~\{b_n\} of real numbers?

A . lim sup n a n = lim sup n a 3 n . B . lim sup n ( a n + b n ) = lim sup n a n + lim sup n b n . C . lim sup n ( a n + 1 a n ) = 0. \begin{aligned} {\rm A.} &&& \limsup_{n\to\infty} a_n = \limsup_{n\to\infty} a_{3n}. \\ {\rm B.} &&& \limsup_{n\to\infty} (a_n+b_n)=\limsup_{n\to\infty} a_n+\limsup_{n\to\infty} b_n. \\ {\rm C.} &&& \limsup_{n\to\infty}(a_{n+1}-a_n)=0. \end{aligned}


This is a part of the College Calc problem set. You can find more problems here .

A \rm A only C \rm C only B , C \rm B,~C only None of them All of them B \rm B only A , C \rm A,~C only A , B \rm A,~B only

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1 solution

Boi (보이)
Dec 24, 2020

A. False

Consider the sequence { a n } : = 3 , 2 , 1 , 3 , 2 , 1 , 3 , 2 , 1 , . \{a_n\}:=3,2,1,3,2,1,3,2,1,\cdots. We have

lim sup n a n = 3 ; lim sup n a 3 n = 1. \limsup_{n\to\infty}a_n = 3;\quad \limsup_{n\to\infty}a_{3n}=1.


B. False

Consider the sequences { a n } : = 1 , 2 , 1 , 2 , 1 , 2 , \{a_n\}:=1,2,1,2,1,2,\cdots and { b n } : = 2 , 1 , 2 , 1 , 2 , 1 , . \{b_n\}:=2,1,2,1,2,1,\cdots. We have

lim sup n a n = lim sup n b n = 2 ; lim sup n ( a n + b n ) = 3. \limsup_{n\to\infty}a_n=\limsup_{n\to\infty}b_n=2;\quad \limsup_{n\to\infty}(a_n+b_n)=3.


C. False

Consider the sequence { a n } : = 0 , 1 , 0 , 1 , 0 , 1 , . \{a_n\}:=0,1,0,1,0,1,\cdots. We have that { a n + 1 a n } = 1 , 1 , 1 , 1 , 1 , 1 , , \{a_{n+1}-a_n\}=1,-1,1,-1,1,-1,\cdots, such that

lim sup n ( a n + 1 a n ) = 1. \limsup_{n\to\infty}(a_{n+1}-a_n)=1.

Oh, I confused bounded with convergent and proved all of them true. Seeing the answer None of them, I was like wait, wtf??!!

Veselin Dimov - 5 months, 2 weeks ago

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Haha well yeah the problem's meant to be confusing

Boi (보이) - 5 months, 2 weeks ago

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