# Finding divergence/convergence by direct comparison test

1. Mar 24, 2012

### OnceKnown

1. The problem statement, all variables and given/known data
$\sum$$^{∞}_{1}$1/n$^{n}$

2. Relevant equations Direct comparison test

3. The attempt at a solution Since the main factor in the equation is the exponent that would be changing as n goes to infinity, I know that from the p series as p > 1 the the series converges. So I know that I would be comparing the original equation to

$\sum$$^{∞}_{1}$1/n$^{2}$

And I know that I need to show:

0 $\leq$ $\sum$$^{∞}_{1}$1/n$^{n}$ $\leq$ $\sum$$^{∞}_{1}$1/n$^{2}$

but I don't know how to show

$\sum$$^{∞}_{1}$1/n$^{n}$ $\leq$ $\sum$$^{∞}_{1}$1/n$^{2}$

mathematically. Would I just blatantly say that the original term is smaller than the p series just by looking at it?

2. Mar 24, 2012

### fzero

The comparison test requires you to show that

$$\left| \frac{1}{n^n} \right| \leq \left| \frac{1}{n^2} \right|$$

for sufficiently large $n$. Try to find nice factors that you can multiply both sides of the equation to compare something to 1.