Is $\sum\frac{|a_n|}{n}$ convergent , while $\sum a_n^2$ is convergent?
I tried from Cauchy definition but failed.
How can it be done some common way?
$\sum a_n^2 $-convergent $\Rightarrow? \sum\frac{|a_n|}{n} -$ convergent
$\sum a_n^2 $-convergent $\Rightarrow? \sum\frac{|a_n|}{n} -$ convergent
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sequences-and-series
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0You may as well assume $a_n$ non-negative – 2017-01-24
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0I assume that you are asking (since this seems the most reasonable interpretation) that $\sum_{n} a_n^2$ converging $\implies \sum_{n} \frac{a_n}{n}$ converges. This follows from the fact that $\sum_{n}2\frac{a_n}{n} \le \sum_{n} a_n^2 + \frac{1}{n^2}$, since the inequality holds termwise for the two series because $(a_n - \frac{1}{n})^2 \ge 0$. – 2017-01-24
1 Answers
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Your wording is ambiguous, presumably you meant iff? If $\sum_n |a_n|^2$ converges then so does $\sum_n \frac{a_n}{n}$ by the Cauchy Schwarz inequality. The reverse direction is not true. Take $a_n=1/\sqrt{n}$.