Since the series $\sum_{n=1}^\infty a_n$ is convergent, so $\lim_{n\to\infty} a_n=0.$

Consider $$\lim_{n\to\infty} {\left(\dfrac{{a_n}^{1/2}}{n}\right)\over{\left(\dfrac{1}{n^2}\right)}} $$ which is equivalent to $\lim_{n\to\infty} n{a_n}^{1/2}$

The above are my steps for solving the question. In the question it is given that infinite series $ \sum a_n$ is convergent, now I am trying to use limit comparison test to conclude the infinite series of $$\frac{{a_n}^{1/2}}{n}$$ is convergent. But I am not sure how to evaluate this limit : $\lim_{n\to\infty} n{a_n}^{1/2}$.
Or anyone has other method to show that the infinite series $$\sum \frac{{a_n}^{1/2}}{n}$$ is convergent given the infinite series $\sum a_n$ is convergent?

  • $\begingroup$ @Nilan is (1/n)^2 $\endgroup$ – UnusualSkill Apr 23 '15 at 8:15
  • 1
    $\begingroup$ I fixed it. Is it correct? $\endgroup$ – Bumblebee Apr 23 '15 at 8:27
  • $\begingroup$ I think Dirichlet test will work for your question. $\endgroup$ – Bumblebee Apr 23 '15 at 10:00

We can use Cauchy-Schwarz inequality. In fact $$\left(\sum_{n\geq1}\frac{a_{n}^{1/2}}{n}\right)^{2}=\left(\lim_{N\rightarrow\infty}\sum_{n\leq N}\frac{a_{n}^{1/2}}{n}\right)^{2}=\lim_{N\rightarrow\infty}\left(\sum_{n\leq N}\frac{a_{n}^{1/2}}{n}\right)^{2}\leq\lim_{N\rightarrow\infty}\sum_{n\leq N}a_{n}\lim_{N\rightarrow\infty}\sum_{n\leq N}\frac{1}{n^{2}}<\infty. $$


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