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This question is motivated by here.

I don't see why the monotone condition is necessarily. Especailly if $a_n \geq 0$. My answer is as follows because $(a_n^2)$ converges, we can find $N : n > N \implies |a_n|^2 < \epsilon \implies |a_n| < \epsilon.$ I am guessing this only proves the absolute convergence and not the convergence of $a_n$.

Lemon
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1 Answers1

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You are correct in saying that monotonicity is not needed if $a_n\ge0$, but if not then you can have an example like $$a_n=(-1)^n$$ in which $a_n^2$ is a constant sequence which converges to $1$, while $a_n$ oscillates between $1$ and $-1$ and does not converge.

Comments.

  • Your proof is OK if the limit claimed is $0$, otherwise you have to start with something involving $|a_n^2-L|<\varepsilon$.
  • "Absolute convergence" applies to series, not to sequences.
David
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  • I borrowed that term from series…I feel like I actually misread the original question completely... – Lemon Sep 24 '14 at 02:37