Iam stuck with this proof. There seems to be no property to help.
If $\sum_1^\infty|a_n|<\infty$ then show that $\sum_1^\infty{a_n^2}<\infty$ and that the reverse isnt true.
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GorillaApe
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Well a well known example that the converse is not true is the sum of inverse of squares and the harmonic sum. – cirpis Jan 23 '15 at 12:08
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The first series converges, so $|a_n|\to0$ as $n\to\infty$. Because of this, there exists an $N$ such that for all $n>N$, $|a_n|<1$. If $|a_n|<1$, then $a_n^2<|a_n|$. Eventually the terms of the new series are less than the terms of the old series. Then by direct comparison, the new series converges.
As for a counterexample to the reverse statement, other answers have already brought up the harmonic and over-harmonic series.
Regret
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My problem is that in case $a_n$ in the beginning before $a_n$<1 ,$a_n^2$ is way larger and produce a bigger sum – GorillaApe Jan 23 '15 at 12:29
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@Parhs: But you see, before $a_n<1$ there exists only a finite number of terms. It does not matter how much larger $a_n^2$ is, as the finite sum of finite terms is finite. – Regret Jan 23 '15 at 12:31
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So direct comparison doesnt necesarily mean something for their sums (comparison) ,interesting .eg $ \sum _{n=1}^{\infty } \left(\frac{100}{n^2}\right)^2 $ > $\sum _{n=1}^{\infty } \left(\frac{1}{n^2}\right)$ . thanks – GorillaApe Jan 23 '15 at 12:36
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@Parhs: It is true that the series with $a_n^2$ can be much larger than the series with $|a_n|$, but both will converge. By "direct comparison" I meant this convergence test: http://en.wikipedia.org/wiki/Direct_comparison_test – Regret Jan 23 '15 at 12:37
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teacher told me that this is poor proof and not valid and that there is simpler one. – GorillaApe Jan 26 '15 at 11:22
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@Parhs: Why is it a poor proof? Of course, it is not a fully rigorous representation, but it should not be difficult at all to generate one from this. – Regret Jan 26 '15 at 14:25
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I dont know he didnt tell me.Maybe he is going to tell me but I am not sure if he will. In case I'll post it. I'll show him this again... – GorillaApe Jan 26 '15 at 14:46
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@Parhs: If you find out what the simpler proof is, or why this one is poor, please let me know in the comments here. – Regret Jan 26 '15 at 22:45
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I am going to ask why it is poor but probably he isnt going to tell me(cant insist he'll get annoyed).However after asking lots of other classmates someone showed me the solution . It is like this http://math.stackexchange.com/a/493782/17446 . – GorillaApe Jan 27 '15 at 09:26
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$\sum_1^\infty{a_n^2} \leq (\sum_1^\infty|a_n|)(\sum_1^\infty|a_n|) < \infty$
A example to show reverse untrue.
See $a_n = \frac {1}{n} $
Brian
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1@Parhs $\forall n , a_n^2 \leq |a_n|(\sum_1^\infty|a_n|) \Rightarrow \sum_1^\infty{a_m^2} \leq (\sum_1^\infty|a_m|)(\sum_1^\infty|a_n|)$ – Brian Jan 23 '15 at 12:31
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did you figured out randomly the $ a_n^2 \leq |a_n|(\sum_1^\infty|a_n|)$? – GorillaApe Jan 26 '15 at 11:40
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Hints: Does $a_n^2\le |a_n|$ eventually hold? What do you know about the harmonic series?
Ittay Weiss
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it doesnt hold .But how do harmonic series relate? they are $1/n^r$ not $1/a_n^r$ – GorillaApe Jan 23 '15 at 12:15
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phd professor told me that this is really poor and unacceptable ... – GorillaApe Jan 26 '15 at 11:37
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@Parhs it was only a hint. The work of turning this into a solution to the problem was left for you. Besides, please avoid leaving meaningless remarks such as "Mr. so and so, who happens to hold a PhD, told me this is really poor, but either did not supply any more criticism, or if any such criticism was supplied I'll just keep it to myself so that I leave a cryptic remark". – Ittay Weiss Jan 26 '15 at 15:32
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Yes it is pointless. However in case I get a solution I'll share it. – GorillaApe Jan 26 '15 at 15:39
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I have solutions which are fine but are not acceptable from professor. (I study computer science) – GorillaApe Jan 26 '15 at 15:43