# For what value of $\alpha$, does this series converge?

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?

• Yes! Forgot to add that, part of the hypothesis – Karl May 7 '18 at 0:02
• $$\sum \frac{\sqrt{a_n}}{n^\alpha} \le (\sum \frac{1}{n^{2\alpha}})^{1/2}(\sum a_n)^{1/2}$$ so $\alpha > .5$ works. – mathworker21 May 7 '18 at 0:12
• Taking $a_n=\frac 1{n^{1+\epsilon}}$ shows that you can't have $\alpha <.5$ ... That just leaves $\alpha =.5$ to sort out. – lulu May 7 '18 at 0:19
• $a_n = \frac{1}{n\log^2 n}$ rules out $\alpha = .5$ (and less than .5). – mathworker21 May 7 '18 at 4:56
• @mathworker21 Would you explain both of your comments?. I don't follow. What is wrong with the reasoning that i used (Root Test). Thanks ! – Karl May 7 '18 at 16:52

$$\textbf{My comments}$$:

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.

.

$$\textbf{Root test}$$:

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 $$\alpha$$).

• i though since $\sum a_{n}$ converge (From the hypothesis of the problem) that $$\lim_{x\to\infty} a_{n} < 1$$ (Root test), is this wrong to asume? And then $$\lim_{n \to \infty} \frac{(a_n)^{1/2n}}{n^{\alpha/n}} <1,$$ because the numerator would be < 1. At this point i know is wrong but i do not know why. Thanks. – Karl May 7 '18 at 18:37
• Antoher thing, why do you consider $a_n = \frac{1}{n\log^2n}$, being a particular sequence.? – Karl May 7 '18 at 18:44
• I have read your answer and i understand the use of the Cauchy-Schwarz inequality, but i dont know why you consider that particular sequence, and why you can do it, it never occurred to me. Thanks – Karl May 7 '18 at 19:13
• @Karl the motivation for choosing $a_n = \frac{1}{n\log^2 n}$ stems from the fact that $\sum \frac{1}{n\log^2 n}$ is summable while $\sum \frac{1}{n\log n}$ is not. And this fact is exactly what I use – mathworker21 May 8 '18 at 1:49
• I was trying to say that your initial question is unclear. Is the sequence $(a_n)_n$ fixed and you're asking which $\alpha$ work for that sequence? Or are you asking which $\alpha$ work for every sequence $(a_n)_n$? For the first question, the answer of course depends on $(a_n)_n$. For the second question (which is the question I answered), I chose that particular sequence $(a_n)_n$ because I knew it is "barely summable" and would thus solve the problem for us – mathworker21 May 8 '18 at 5:08