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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?

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

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 $\alpha$).

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  • $\begingroup$ 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. $\endgroup$ – Karl May 7 '18 at 18:37
  • $\begingroup$ Antoher thing, why do you consider $a_n = \frac{1}{n\log^2n}$, being a particular sequence.? $\endgroup$ – Karl May 7 '18 at 18:44
  • $\begingroup$ 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 $\endgroup$ – Karl May 7 '18 at 19:13
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    $\begingroup$ @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 $\endgroup$ – mathworker21 May 8 '18 at 1:49
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    $\begingroup$ 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 $\endgroup$ – mathworker21 May 8 '18 at 5:08

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