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Suppose $(a_n)$ is a monotone sequence. Prove that if $(a_n^2)$ is convergent, then so is $(a_n)$.

How do I use the monotone convergence theorem to prove this?

M. Vinay
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user178358
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2 Answers2

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Since $(a_n)$ is monotone, it suffices to prove that $(a_n)$ is bounded.

Since $(a_n^2)$ is convergent, it is bounded. Therefore, there exists some $M > 0$ such that $|a_n^2| \leq M$ for all $n$. But then $|a_n| \leq \sqrt{M}$ for all $n$, so $(a_n)$ is bounded, as desired.

  • Doesn't $(a_n^2)$ being convergent means for $n > N \implies$ $|a_n|^2 < \epsilon$ for all $\epsilon$? So $n > N \implies |a_n| < \sqrt{\epsilon}$, meaning it is convergent? Why is the monotone condition necessary? – Lemon Sep 24 '14 at 02:18
  • @sidht: What you have written shows that if $a_n^2$ converges to zero then $a_n$ also converges to zero. This is true. However, convergence of $a_n^2$ to some other limit does not necessarily imply convergence of $a_n$. Consider for example $a_n = (-1)^n$. Then $a_n^2 = 1$ for all $n$, so certainly $a_n^2$ converges to $1$. But $a_n$ does not converge. What we can say is that if $a_n \geq 0$ for all $n$ (or all sufficiently large $n$), and $a_n^2 \rightarrow L$, then $a_n \rightarrow \sqrt{L}$. This is true because the square root function is continuous. –  Sep 24 '14 at 02:48
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If a sequence converges it is bounded. Since $|a_n^2| = |a_n|^2 \le M$ for all $n$ it follows that $-\sqrt{M} \le a_n \le M$. Since $\{a_n\}$ is monotone and bounded, you have the hypothesis of the MCT.

David P
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