Given the limit to the infinity of sequence $A_n = 0$. Does the series $\Sigma_{n=1}^{\infty} A_n^2$ have to be convergent or divergent? If it is one of them, prove it. If it is either, give an example.
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- $\begingroup$ Are you familiar with the p series? $\endgroup$Ian– Ian2016-03-20 02:19:45 +00:00Commented Mar 20, 2016 at 2:19
- $\begingroup$ yes I am familiar @Ian $\endgroup$London Yoil– London Yoil2016-03-20 02:21:15 +00:00Commented Mar 20, 2016 at 2:21
- $\begingroup$ Then try $A_n=n^p$ for various negative values of $p$. $\endgroup$Ian– Ian2016-03-20 02:21:40 +00:00Commented Mar 20, 2016 at 2:21
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1 Answer
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I change the notations of you and show the sequence with $a_n$ and series with $S=\Sigma (a_n)^2$.
If $$\lim_{n\to\infty} a_n=0$$
then
$$\Sigma a_n^2$$
can be either convergence or divergence.
Convergence example:
$a_n=\frac1n , ~~~~~~~~~~~~ S=\Sigma a_n^2=1+\frac14+\frac19+\dots=\text{convergent}$
Divergence example:
$a_n=\frac1{\sqrt n} , ~~~~~~~~~~~~ S=\Sigma a_n^2=1+\frac12+\frac13+\dots=\text{divergent}$