Rudin chap 3 exercise 7.

Question

The question is basically to prove that convergence of $\sum{a_n}$ implies convergence of $\sum {\frac{\sqrt a_n}n}$ with $a_n>0.$

Basically what I tried to show is that:

Since $\sum{a_n}$ converges, then $\lim a_n= 0$ when $n \rightarrow \infty$ , and so $\sqrt a_n \rightarrow 1$ when $n \rightarrow \infty,$ which means it is bounded. On the other hand, $\frac1n \rightarrow 0$ as $n \rightarrow \infty$ and is always decreasing. Per theorem 3.42 in Rudin, that means $\sum{\frac{\sqrt a_n}n}$ also converges.

Is that correct or am I missing something? The solutions on the net appear to differ.

Answer 1

• If $\lim_{n \to \infty} a_n = 0$, then $\lim_{n \to \infty}\sqrt{a_n}=0$.

• To use theorem $3.42$ , which is the Dirichlet's test, you need to show that $\sum_n \sqrt{a_n}$ is bounded, of which it is not known to be true.

Possible direction for the question:

• consider Cauchy-schwarz.

Answer 2

Cauchy-Schwarz:

\begin{align*} \sum\dfrac{\sqrt{a_{n}}}{n}\leq\left(\sum a_{n}\right)^{1/2}\left(\sum\dfrac{1}{n^{2}}\right)^{1/2}<\infty. \end{align*}