My professor worked this following problem in class, and I'm having some difficulty following the logic.
We had the alternating series, $\sum\limits_{n=2}^{\infty} \frac{(-1)^n \sqrt{n}}{\ln n}$. This fails the alternating series test, as $\lim\limits_{n \to \infty} \frac{\sqrt{n}}{\ln n} = \infty$. He used this as a basis to say that, by the Divergence Test, the series diverges.
I can't follow this, though. The Divergence Test, if I'm not mistaken, is on the entirety of the general term of the series, $\frac{(-1)^n \sqrt{n}}{\ln n}$. The test that he ran should show that the series is not absolutely convergent, though it could still be conditionally convergent. The other problem is that $\lim\limits_{n \to \infty} \frac{(-1)^n \sqrt{n}}{\ln n}$ is not zero because it doesn't exist; the sign alters back and forth. But wouldn't this logic apply to any alternating series, for example, $\sum\limits_{n=1}^{\infty} (-1)^n \sin^{-1} \frac{1}{n}$?
Revision to the above: After running this new series through wolframalpha, it does indeed some to have a limit of zero, though I can't seem to make out why, as $(-1)^n$ doesn't converge, and if we break up the limit of a product into the product of the limits, we get one term that doesn't converge (oscillates between $1$ and $-1$) and one term that goes to $0$. Does the $0$ "win out"?
More compactly, my questions are:
(a) Is it sufficient to run the alternating series test on the absolute value of the general term of the series? Or should we run it on the entire alternating series, and conclude that the oscillation of the $(-1)^n$ factor will prevent the limit from converging?
(b) Is it true that the lim as $n \to \infty$ of the entirety of the general term will always be zero for a series that converges by the alternating-series test? Is this true because the $0$ factor "wins out" against the oscillating factor?
Thanks in advance.