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Ok, so this one stumped me completely.

Determine for which values of $\alpha$ the following sum converges: $$\sum_{n=1}^{\infty} \frac{n}{\sqrt{4+n^\alpha}}$$

My gut instinct was that for any $\alpha <2$ it can not converge, but a little bit of fiddling with WolframAlpha showed me that it actually does not converge for $\alpha \leq \approx 4$.

I couldn't find the exact figure, and even if I did, I'm not sure HOW I would go about it. I tried the integral test but it became much more complex than I know how to deal with, I tried the limit test but again ran into complications...

I get the feeling that before I can even begin to answer this question I need to simplify the expression so that I can isolate $n^\alpha$ but I'm not sure how to go about it.

Any tips would be appreciated!

Thanks in advance.

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hint: $$\sum_{n=1}^{\infty} \frac{n}{\sqrt{4+n^\alpha}}<\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^\alpha}}=\sum_{n=1}^{\infty} n^{1-\frac{\alpha}{2}}$$ now applying well known fact that $$\sum_{n=1}^{\infty}n^x$$ converges when $x<-1$ thus $$1-\frac{\alpha}{2}<-1$$ $$\alpha>4$$

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  • $\begingroup$ Thanks, very well explained. Cheers! :-) $\endgroup$ – Elad Avron Jun 27 '16 at 10:24
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Hint. One may observe that, as $n \to \infty$, we have $$ \frac{n}{\sqrt{4+n^\alpha}}=\frac1{n^{\alpha/2-1}}\frac1{\sqrt{1+4/n^\alpha}} \sim \frac1{n^{\alpha/2-1}} $$ which gives a convergent series if and only if $\alpha/2-1>1$ that is if and only if $\alpha>4$. Then this is the case for the initial series by using the comparison test.

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  • $\begingroup$ Thank you, it's a good direction but eventually @avz2611 's solution (which is basically the same, admittedly) rang the bell in my head a bit faster. $\endgroup$ – Elad Avron Jun 27 '16 at 10:23

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