1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ it's same as $\frac{1}{n^{2/3}}$, so no, it doesn't converge $\endgroup$ – mathworker21 Dec 12 '18 at 5:36
4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

We have that

$$\frac{n}{\sqrt[3]{8n^5-1}}\sim \frac1{2n^{2/3}}$$

therefore we can conclude that the given series diverges by limit comparison test with $\sum \frac1{n^{2/3}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.