Finding divergence/convergence by direct comparison test

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OnceKnown
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Homework Statement


[itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex]


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

[itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

And I know that I need to show:

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

but I don't know how to show

[itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex] [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

mathematically. Would I just blatantly say that the original term is smaller than the p series just by looking at it?
 
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OnceKnown said:
And I know that I need to show:

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

The comparison test requires you to show that

[tex]\left| \frac{1}{n^n} \right| \leq \left| \frac{1}{n^2} \right|[/tex]

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