Does the denominator become larger faster

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Homework Statement


Determine if convergent or divergent. Determine limit if convergent.

Homework Equations


a_{n} = \frac{n!}{n^n}

The Attempt at a Solution


As per the hint, i use 1/n to compare.

however, how is this statement true:

\lim_{x\to\infty} \frac{n!}{n^n} <= \lim_{x\to\infty} \frac{1}{n}??

does the denominator become larger faster than the numerator making it a smaller number than the 1/n?
 
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whatlifeforme said:

Homework Statement


Determine if convergent or divergent. Determine limit if convergent.

Homework Equations


a_{n} = \frac{n!}{n^n}

The Attempt at a Solution


As per the hint, i use 1/n to compare.

however, how is this statement true:

\lim_{x\to\infty} \frac{n!}{n^n} <= \lim_{x\to\infty} \frac{1}{n}??

does the denominator become larger faster than the numerator making it a smaller number than the 1/n?

Think about it. E.g. 3!/3^3 is less than 1/3. Why is that? That's (1*2*3)/(3*3*3).
 
dick said:
think about it. E.g. 3!/3^3 is less than 1/3. Why is that? That's (1*2*3)/(3*3*3).

update: nevermind.
 
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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