Question

Exercise 2.2.5. Let [[x]] be the greatest integer less than or equal to x. For example, [[π]] = 3 and [[3]] = 3. For each sequence, find lim an and verify it with the definition of convergence. (a) an = [[5/n]], (b) an = [[(12 + 4n)/3n]].

Answer 1

Answer:

a) $\lim_{n\rightarrow \infty}a_n=0$

b) The limit is [[4/3]]=1, that is, $\lim_{n\rightarrow \infty}a_n=1$

Step-by-step explanation:

The definition of the limit of a sequence (an) is as follows:

$\lim_{n\rightarrow \infty}a_n=L$ (the sequence (an) converges to L) means that given $\epsilon>0$, there exists some natural number N such that if n≥N, then $|a_n-L|$

We will use that [[x]] is a increasing function. That is, if x≤y then [[x]]≤[[y]]

a) For all $\epsilon>0$, take N≥6. Then 1/N ≤1/6 and 5/N ≤ 5/6 < 1. We have that 0<5/N<1, therefore [[5/N]]=0 (0 is the biggest integer less than or equal than x, for any x on the interval 0<x<1)

For n≥N, we have that 1/n ≤ 1/N, hence 0< 5/n ≤ 5/N < 1 . As we discussed above, [[5/n]]=0. Then |a_n|=0, therefore $|a_n-0|=|a_n|=0$  

b) For all $\epsilon>0$, take N≥3/4 (any N will do). Then 1/N ≤ 4/3 and 1/4N ≤ 1/3. Note that a_n=[[(12 + 4n)/3n]]=[[1/4n + 4/3]]

For n≥N, we have that 1/4n ≤ 1/4N, hence 1< 4/3< 1/4n +4/3 ≤ 1/4N+4/3 ≤ 5/3 <2.If 1<x<2 then [[x]]=1 , hence a_n=[[1/4n +4/3]]=1  

Then |a_n-1|=|1-1|=0, therefore $|a_n-1|=0$