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
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calculus
real-analysis
sequences-and-series
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0I mean less than infinity or what is the same that the series converge – 2012-10-09
1 Answers
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Try $a_n = \frac{1}{(n+1)\log(n+1)}.$
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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