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I have the following series :

$$\sum_{k=1}^{\infty }\left ( \frac{k}{\sqrt{4k^{3}+1}} \right )$$

And I am trying to see if it's convergent or divergent. I first thought about the integral test, but it looks like it will not be easy to integrate it. I found a hint that it can be solved using the comparison test with $$b_{n}=\frac{1}{k}$$, but I don't understand where does $$\frac{1}{k}$$ come from.

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    $\begingroup$ $\sqrt{k^3 + 1} \approx \sqrt{k^3} = k^{3/2}$, so $$\frac{k}{\sqrt{k^3 + 1}} \approx \frac 1 {\sqrt k} > \frac 1 k$$ $\endgroup$ – user296602 Jul 18 '16 at 3:53
  • $\begingroup$ If the $\approx$ in the above comment makes you uneasy, observe that for any positive $k$, we have $$\begin{aligned}\frac{k}{\sqrt{k^3 + 1}} \geq \frac{1}{2\sqrt{k}}&\iff 2k^{3/2} \geq \sqrt{k^3 + 1} \\&\iff 4k^3 \geq k^3 + 1 \\&\iff k^3 \geq \frac{1}{3} \\ \end{aligned}$$ As the last inequality is certainly true for all $k \geq 1$, so is the first, so the series diverges by comparison with $\sum_{k=1}^{\infty}1/(2\sqrt{k})$. $\endgroup$ – Bungo Jul 18 '16 at 4:14
  • $\begingroup$ by euler mac laurin the the partial sums of the series behave as $\sqrt{N}+C+\mathcal{O}\left(\frac{1}{N^{5/2}}\right)$ which diverges $\endgroup$ – tired Jul 18 '16 at 23:08
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you can find out the limit to compare can be computed by the highest degree coeff. in numerator divided by the highest degree coeff. in numerator. in your case it will be $$\sum \frac{k}{\sqrt{4k^3 +1}}$$

taking square and it will be

$$\sum \frac{k^2}{4k^3+1}$$

now we can write the limit to compare

$$\frac{k^2}{k^3}=\frac{1}{k}$$

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$$\sum_{k\geq1}\frac{k}{\sqrt{k^{3}+1}}\geq\frac{1}{\sqrt{2}}\sum_{k\geq1}\frac{k}{\sqrt{k^{3}}}=\frac{1}{\sqrt{2}}\sum_{k\geq1}\frac{1}{\sqrt{k}}\geq\frac{1}{\sqrt{2}}\int_{1}^{\infty}\frac{1}{\sqrt{x}}dx=\infty.$$

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