Question

Give an example of a sequence satisfying the condition or explain why no such sequence exists. (a) A divergent sequence {an} such that {a2n} converges. (b) A monotonically increasing sequence that converges to 0. (c) A convergent sequence that is not bounded. (d) A monotonically decreasing bounded sequence that diverges.

Answer 1

Answer:

Step-by-step explanation:

a) Consider the sequence $a_n =1$ if n is odd, and $a_n= -1$ if n is even. So, the sequence diverges (since as n tends to infinity the sequence doesn't approach any particular number), but the subsequence of the even integers is convergent to -1 since it is constant.

b) consider the sequence $a_n = -e^{-n}$. The function f(x) = $e^{-x}$ when x is real is a monotolically decreasing function and tends to 0. Then, when multiplying by a minus sign, it becomes a monotonically increasing function that tends to 0. Hence, the given sequence is monotonically increasing and converges to 0.

c) Suppose that the sequence $a_n$ converges to a. So, from an specific n and on, the values of $a_n$ are really close to a. So, for almost all the value of the sequence, they are less than a+1 and greater than a-1. Hence it must be bounded.

d) It is a theorem that a monotonically decreasing/increasing sequence that is bounded must converge, so such a sequence can't exist.