# 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??

• You are doing it wrong. Have a look here. If you are annoyed by the +1 in $n+1$, just take $(n+1)$-th root (or, equivalently, translate your indices by 1) – Jean-Claude Arbaut Nov 28 '13 at 23:54
• Your reasoning is not correct. You must be careful with sums of products, as they are not in general equal to products of sums. That is, it is not always the case that $\sum a_{n}b_{n} = \sum a_{n} \sum b_{n}$ (in fact, this is quite rare!) – Alex Wertheim Nov 28 '13 at 23:55

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?

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.

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.