Series: $\Sigma n!/n^n$ - Why Does it Converge?

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Homework Statement


\Sigman!/n^n

index n=1 to infinity


Homework Equations





The Attempt at a Solution


Using the Ratio test (limit as n goes to infinity of a_{n+1}/a_{n})
and found that the series converges.

However, I thought that factorials grew faster than exponential functions. Therefore, it would diverge, right?

Could someone explain why? Did I just do something wrong?
 
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Exponents of the form n^n grows much faster than factorials of the form (n!) because the factorial is a multiplication of n terms, the majority of which are less than n, and the power is a multiplication of n terms, all of which are equal to n.
 
Factorials don't grow faster than exponentials of the sort you're working with. Just think about it: n! = 1 * 2 * 3 * ... * n. You have n factors, of which the largest is n.
n^n = n * n * n * ... * n. Here you have n factors, all of which are n. Clearly this exponental is larger than the factorial above.
 
Exponentials with a fixed base, like e^n or 2^n, grow more slowly than n!.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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