Suppose $\{x_n\}$ is a monotonically increasing sequence and the subsequence $\{x_{2n}\}$ is convergent. Justify $\{x_n\}$ is convergent.

Question

I think I did this correctly:

Since $\{x_{2n}\}$ is convergent, it is bounded. Since $\{x_{2n}\}$ is a bounded subsequence of $\{x_{n}\}$, $\{x_{n}\}$ is also bounded. Hence, $\{x_{n}\}$ is monotone and bounded. Therefore, $\{x_{n}\}$ is convergent by the monotone convergence theorem.

I'm just not sure about this statement: "Since $\{x_{2n}\}$ is a bounded subsequence of $\{x_{n}\}$, $\{x_{n}\}$ is also bounded."

You're right to be not sure. As it stands, it doesn't imply the boundedness of ${x_n}$. You must use the monotonicity for the conclusion. — Daniel Fischer, Oct 12 '13 at 21:19

score 3 · Answer 1 · edited Oct 13 '13 at 00:34

Yes, you are correct: Suppose that $M$ is a bound on the sequence $\{x_{2n}\}$. Now if we consider an odd number $2k - 1$, then we can say

$$x_{2k - 1} \le x_{2k} \le M$$

by monotonicity. Hence $M$ is a bound on both the even and odd parts of the sequence, thus on the sequence as a whole.

But it's not true in general that a sequence with a bounded subsequence is bounded; take

$$x_n = \left\{\begin{array}{c} 0 & : n \text{ is even} \\ n & : n \text{ is odd} \end{array} \right.$$

score 1 · Answer 2 · answered Oct 13 '13 at 01:27

1

Your hypotheses can be weakened. Instead of using $\{x_{2n}\}$, you can use $\{x_{n_i}\}$, where $(n_i)_{i=1}^{\infty}$ is any strictly increasing sequence of positive integers (such as $2_i$, $i^2$, or $2^{2^{...^2}}$ with $i$ levels of exponents).

answered Oct 13 '13 at 01:27

marty cohen

107,799

Suppose $\{x_n\}$ is a monotonically increasing sequence and the subsequence $\{x_{2n}\}$ is convergent. Justify $\{x_n\}$ is convergent.

2 Answers2