I Have the next Series and i need to find for what value of $\alpha$ the series converge.
$\sum \frac{\sqrt{a_{n}}}{n^{\alpha}}$
Hipothesis: $\sum a_{n}$ converge, $a_{n}$ positive
So, waht i tried to use is the "Root Test" since i know that $$\lim_{n\to\infty} \sqrt[n]{a_{n}} < 1$$ (Hypothesis from $\sum a_{n}$ converge).
Doing that ends on the series converging for all $\alpha$. I sense there is something wrong. Any advice on how to work with this series. Why is neccesary the Hypothesis?
$\textbf{My comments}$:
The Cauchy-Schwarz inequality yields $$\sum \frac{\sqrt{a_n}}{n^\alpha} \le (\sum \frac{1}{n^{2\alpha}})^{1/2}(\sum a_n)^{1/2},$$ so if $\alpha > \frac{1}{2}$, the series converges. Now consider $a_n = \frac{1}{n\log^2n}$. Indeed, $\sum a_n < \infty$ (by integral test for example), but $\sum \frac{\sqrt{a_n}}{n^\alpha} = \sum \frac{1}{n^{\frac{1}{2}+\alpha}\log n}$ so if $\alpha \le \frac{1}{2}$, this series diverges.
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$\textbf{Root test}$:
Note $$\lim_{n \to \infty} (\frac{\sqrt{a_n}}{n^\alpha})^{1/n} = \lim_{n \to \infty} \frac{(a_n)^{1/2n}}{n^{\alpha/n}} = 1,$$ so the root test doesn't tell us anything (for any $\alpha$).