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.
$\endgroup$ 32 Answers
$\begingroup$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
$$\frac{k^2}{k^3}=\frac{1}{k}$$
$\endgroup$ $\begingroup$$$\sum_{k\geq1}\frac{k}{\sqrt{k^{3}+1}}\geq\frac{1}{\sqrt{2}}\sum_{k\geq1}\frac{k}{\sqrt{k^{3}}}=\frac{1}{\sqrt{2}}\sum_{k\geq1}\frac{1}{\sqrt{k}}\geq\frac{1}{\sqrt{2}}\int_{1}^{\infty}\frac{1}{\sqrt{x}}dx=\infty.$$
$\endgroup$
