# Does the series $\sum_{n=1}^\infty \frac{n}{\sqrt[3]{8n^5-1}}$ Converge?

$$\sum_{n=1}^\infty \frac{n}{\sqrt[3]{8n^5-1}}$$

From the tests that I know of:

Divergence Test: The limit is ≠ to a constant, so inconclusive.

Geometric series: I don't think this could be written in that manner.

Comparison Test/Lim Comparison: Compare to $$\frac{n}{8n^{\frac{5}{3}}}$$

Integral Test: I can't think of a integration method that would work here.

Alternating Series/Root Test don't apply.

Ratio Test: The limit is 1 so inconclusive.

Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?

• it's same as $\frac{1}{n^{2/3}}$, so no, it doesn't converge – mathworker21 Dec 12 '18 at 5:36

Use comparison test,$$\frac{n}{\sqrt[3]{8n^5-1}} \ge \frac{n}{\sqrt[3]{8n^5}} = \frac{1}{2n^{\frac53-1}}=\frac{1}{2n^{\frac23}}$$
Now, use $$p$$-series to make conclusion that it diverges.
$$\frac{n}{\sqrt[3]{8n^5-1}}\sim \frac1{2n^{2/3}}$$
therefore we can conclude that the given series diverges by limit comparison test with $$\sum \frac1{n^{2/3}}$$.