Prove that, if $\left\{a_n\right\}_{n=1}^{\infty}$ converges to A, then $\left\{|a_n|\right\}_{n=1}^{\infty}$ converges to |A|. Is the converse true?
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what have you done so far ? – Tlön Uqbar Orbis Tertius May 04 '15 at 07:23
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Well, I says, since $\left{a_n\right}_{n=1}^{\infty}\to A$, then for ever $\epsilon > 0$, there is $N\in J$ such that $n\geq N$ implies $|a_n - A|<\epsilon$. – Indigo May 04 '15 at 07:28
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Consider $a_n=(-1)^n$, then $|a_n|=1$, hence convergent but $a_n$ is oscillating.
Anurag A
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I'd upvote you both but I can't yet. :/ Thank you for the counter example on the converse statement. – Indigo May 04 '15 at 07:36
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Didn't downvote myself, but I think it's getting downvoted because it answers the half of what he asks that is "hidden" at the end. – Nescio May 04 '15 at 07:39
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No. Try $a_n=(-1)^n $. $\ \ \ \ \ \ \ $
Martin Argerami
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I didn't downvote it. I don't have enough reputation to do so! But it only answers the question of "Is the converse true" and doesn't address the proof, so I think that's why? (It's still helpful, I wouldn't have downvoted it...) – Indigo May 04 '15 at 07:34
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tip: $$||a_n|-|A||\leq|a_n-A|$$
so if $|a_n-A|<\epsilon$...
The converse was already given
Nescio
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1Ooh. So...
Since $\left{ a_n \right}$ converges to A, we have, for every $\epsilon > 0$, there is $N\in J$ such that for $n\geq N$, we have $|a_n - A| < \epsilon$. Since $||a_n| - |A|| \leq |a_n - A| < \epsilon$...... We have that $\left{|a_n|\right}\to |A|$. (...?)
– Indigo May 04 '15 at 07:31 -
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Thanks! I knew it had to be something pretty obvious, and probably to do with that inequality. I have no idea why inequalities always mess me up. – Indigo May 04 '15 at 07:38