if $\sum_{i=1}^n a_i$ converges but $\sum_{i=1}^n (a_i)^2$ diverges, does that mean that $\sum_{i=1}^n a_i$ is conditionally

Question

I don't think this is true, but cannot find an example to disproof it. Either that or i need to prove that $\sum_{i=1}^n |a_i|$ diverges, which I am unclear how to approach the proof.

To get the sum of the $a_n$ terms to converge, you must have $a_n\to 0$. How then does $|a_n|$ compare with $(a_n)^2$ whose sum is hypothesized to diverge? — Oscar Lanzi, Nov 19 '20 at 13:57
Minor notation quibble: $\sum_{i=1}^n$ denotes a finite sum. You want $\sum_{i=1}^\infty$. — Barry Cipra, Nov 19 '20 at 14:12

score 3 · Accepted Answer · edited Nov 19 '20 at 14:08

3

Yes, it is true. That's so because, if $\sum_{n=1}^\infty a_n$ was absolutely convergent, then we would have $\lim_{n\to\infty}a_n=0$ and therefore $a_n^{\,2}\leqslant |a_n|$ if $n$ is large enough. So, by the comparison test (this is where absolute convergence is used), $\sum_{n=1}^\infty a_n^{\,2}$ would converge too.

edited Nov 19 '20 at 14:08

Oscar Lanzi

39,403

answered Nov 19 '20 at 13:58

José Carlos Santos

427,504

1

Yup, you blew my cover. I do hope the OP was able to figure out my comment beforehand. – Oscar Lanzi Nov 19 '20 at 14:01
@OscarLanzi Sorry about that. I only saw your comment after posting my answer. – José Carlos Santos Nov 19 '20 at 14:03
It should be $a_n^2 \le |a_n|$, shouldn't it? – Martin R Nov 19 '20 at 14:05
Corrected, now. – Oscar Lanzi Nov 19 '20 at 14:08
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@OscarLanzi Thank you. – José Carlos Santos Nov 19 '20 at 14:15
I didn't have this approach in mind but its brilliant, Thank you for the help! – unis118 Nov 19 '20 at 14:53

if $\sum_{i=1}^n a_i$ converges but $\sum_{i=1}^n (a_i)^2$ diverges, does that mean that $\sum_{i=1}^n a_i$ is conditionally

1 Answers1