How the convergence of $\sum_{n=1}^\infty \sqrt{n}a_n$ implies the convergence of $\sum_{n=1}^\infty a_n$?
I know this if $a_n>0$. But for arbitrary $a_n$, I have no idea...
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@Albert How does that answer the question, though? – Clement C. Dec 29 '15 at 13:13
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2But $a_n \to 0$ does not imply the convergence of $\sum a_n$ – Jimmy R. Dec 29 '15 at 13:13
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Use the ratio test? – vonbrand Dec 29 '15 at 13:18
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Follows from Abel's theorem : If $\sum c_n$ converges and $\{d_n\}$ is monotone and bounded then $\sum c_nd_n$ converges. Let $c_n=\sqrt{n}a_n$ and $d_n=\frac{1}{\sqrt{n}}$
Rene Schipperus
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I think the statement is the other way round: If every finite sum (i.e. $N$) we have $|\sum_{1\le n \le N} c_n|<M$ for some $M \in \mathbb{R}^+$ and $(d_n)$ is monotone and has $0$ as a limit then $ \sum c_n d_n$ converges. – Marko Karbevski Dec 29 '15 at 13:40
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It seems a situation for Abel's test, quite useful in this non absolutely convergent series case:
- The series $c_n = \sqrt{n}a_n $ yields a convergent sum,
- $b_n = \frac{1}{\sqrt{n}}$ is monotone and bounded.
Then $\sum_\limits{n} b_n c_n = \sum_\limits{n} a_n $ is convergent too.
Laurent Duval
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:I wanted to do this problem just by using the necessary condition for the convergence of series i.e., $\lim_{n\to \infty} c_n$=0.Since $\sum_{n} c_n$ is convergent,this means that $a_n$ be some decreasing sequence.Now in order to show that $a_n$ is convergent sequence in $\mathbb R$ so $a_n$ should be a cauchy sequnce.How can i show that $a_n$ is a cauchy sequence? – Styles Jul 22 '17 at 09:18
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