$\sum_{n=2}^\infty {(-2)^n \over n} $ How does this converge or diverge using the alternate series test? $$\sum_{n=2}^\infty {(-2)^n \over n} $$
When I took the limit I got -2, I also tried using ratio and root test and got the same answer. The answer is supposed to be divergent I think but I thought if Rho<1 it is convergent, unless you can't have negative limits?
 A: The ratio test states that if
$$ \lim\limits_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| =r $$
And $r \gt 1$, then the series diverges. So now we have
$$ \lim\limits_{n\to\infty} \left|\frac{\frac{(-2)^{n+1}}{n+1}}{\frac{(-2)^n}{n}}\right| = \lim\limits_{n\to\infty} \left|\frac{(-2)^{n+1}n}{(-2)^n(n+1)}\right| $$
$$= \lim\limits_{n\to\infty} \left|\frac{(-2)^{n}(-2)n}{(-2)^n(n+1)}\right| = \lim\limits_{n\to\infty} \left|\frac{-2n}{n+1}\right| $$
$$= \lim\limits_{n\to\infty} \left|\frac{2n}{n+1}\right| = \lim\limits_{n\to\infty} \frac{2}{1+\frac1n} =2$$
Therefore by the ratio test,
$$ \sum\limits_{n=2}^{\infty} \frac{(-2)^n}{n}\Rightarrow \mbox{diverges} $$
A: Recall that alternate series test says that
$$\sum_{n=1}^{\infty} (-1)^n a_n$$
converges if $a_n$ is an eventually decreasing sequence to $0$. In your case, $a_n = \dfrac{2^n}n$. This is increasing and diverges. Hence, alternate series test is not applicable in this case.
You can directly conclude that the series doesn't converge, since a necessary condition for a series to converge is that the $n^{th}$ term converges to zero. In your case, the $n^{th}$ terms doesn't converge at all in the first place.
