Finding divergence/convergence by direct comparison test

[OnceKnown]

Homework Statement

\sum^{∞}_{1}1/n^{n}

Homework Equations

Direct comparison test

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?

 

[fzero]

And I know that I need to show:

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

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.