Convergence using Root Test Problem: test if the series converges$$\sum_{n=1}^ \infty \frac {(-2)^{n+1}} {n^{n+1}} $$
My approach:
I see it is equal to $$\sum_{n=1}^ \infty \frac {(-2)^n} {n^n} \cdot \frac {-2} n$$, and $\sum_{n=1}^ \infty \frac {(-2)^n} {n^n}$ converges absolutely using root test, and $\sum_{n=1}^ \infty \frac {-2} n $ diverges by using p-series test. 
So is the original series divergent because convergent * divergent = divergent?
Is convergent * convergent = convergent??
 A: HINT:
$$ \sum_{n=1}^{\infty} \frac{(-2)^{n+1}}{n^{n+1}} = \sum_{n=2}^{\infty} \frac{(-2)^n}{(n-1)^n} $$
Now what does the root test say?
A: If you use the root test, then you need to find the $\limsup$ of

$$ |a_n|^{1/n} = \left( \frac{2}{n} \right)^{\frac{n+1}{n}},$$

as $n\to \infty.$ Try to work out this limit.
A: Root Test 
$$\lim_{x \to \infty} |a_n|^\frac{1}{n}= L$$
If $L < 1$, the series converges absolutely. If $L > 1$, the series diverges. If $L = 1$, the test is inconclusive.
$$a_n = \frac{{(-2)}^{n+1}}{n^{n+1}}$$
$$|\frac{{(-2)}^{n+1}}{n^{n+1}}|^\frac{1}{n} =\frac{{(-2)}^{(n+1) \cdot \frac{1}{n}}}{n^{(n+1) \cdot \frac{1}{n}}} = \frac{(-2)^{(1 + \frac{1}{n})}}{n^{(1 + \frac{1}{n})}}$$
$$\lim_{n \to \infty} \frac{(-2)^{(1 + \frac{1}{n})}}{n^{(1 + \frac{1}{n})}}$$
$\frac{1}{n} \to 0$ as $n \to \infty$
$$\lim_{n \to \infty} \frac{(-2)^{(1 + 0)}}{n^{(1 + 0)}} = \lim_{n \to \infty} -\frac{2}{n} = -\frac{2}{\infty} \to 0$$
Since $L = 0 <1$, the series converges absolutely $\implies$ the series converges in the usual sense.
