Convergence or divergence of alternating series $\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^nn^2+3}{n\sqrt{9n^2+2}}$

Question

Trying to figure out this problem. Alternating series test doesn't seem to work because finding the limit seems to be an endless loop of L'Hopital's rule in the denominator. Ratio test is inconclusive because it turns out to be 1. Comparing seems difficult because of the same. Not to mention difficulty finding something that is less than or greater than the series for all values for comparison. I'm a bit stumped. I suppose that technically, the positive series convergence would be fine as well, as it would prove absolute convergence, but I can't even figure that out.

Answer 1

Hint: The general term does not tend to $0$ so the series is not convergent.

Answer 2

To avoid the infinite loop of L'Hôpital, try writing $$\frac{n^2+3}{n\sqrt{9n^2+2}}=\sqrt{\frac{n^4+6n^2+9}{9n^4+2n^2}}$$

Then considering the limit as $n$ tends to infinity of the expression inside the square root, we see that this tends to $\sqrt{\frac{1}{9}}=\frac{1}{3}$. This is non-zero, so the series diverges.

Answer 3

Your general term is $$ \frac{(-1)^n n^2 + 3}{n \sqrt{9n^2 + 2}} = \frac{n^2}{n^2} \cdot \frac{(-1)^n+\frac{3}{n^2}}{\sqrt{9 + \frac{2}{n^2}}} \text{.} $$ It should be fairly clear that the terms of your series do not decrease to zero, so (first) you cannot apply the alternating series test and (second) the series does not converge because (for both) the limit of the general term is not zero.