Let $a_n>0$ for all $n\in\mathbb{N}$ and the series $\sum\limits_{n=1}^\infty (n+1) a_n^2$ converges. Does it imply the series $$ \sum\limits_{n=1}^\infty a_n $$ is also converges?
Convergence test for positive series
1
$\begingroup$
calculus
real-analysis
sequences-and-series
-
0What do you mean by "$<\infty$"? – 2012-10-09
-
1@Jacob That it converges to a finite number. – 2012-10-09
-
0I mean less than infinity or what is the same that the series converge – 2012-10-09
1 Answers
1
Try $$a_n = \frac{1}{(n+1)\log(n+1)}.$$
-
0lol, beat me by a couple seconds... Will delete mine... – 2012-10-09
-
0mmm. I really don't know what to do with that – 2012-10-09
-
1@JORGEBARRERA: they are suggesting you that you prove that for such $a_n$ you have $\sum(n+1)a_n^2<\infty$ and $\sum a_n=\infty$. – 2012-10-09
-
0@JORGE BARRERA Comparison test shows the first converges and the same test shows the second diverges, hence your hypothesis do not imply $\sum_{i=1}^\infty a_n$ converges – 2013-02-07