Power Series- radius of convergence

Roni1985
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Homework Statement


determine the radius of convergence of the given power series

\sum^{inf}_{n=1}\frac{n!x^n}{n^n}

Homework Equations





The Attempt at a Solution


I did the ratio test
then I had to take the 'ln'
but, my answer is this
|e|<1 for the series to converge.
It never happens but according to the answers the radius is 'e'.
 
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oh I think I got it, I didn't have to lhopital lnx in the middle ...
:\
 
Applying the Ratio Test, we have the sequence

\frac{(n+1)n^n}{(n+1)^{(n+1)}} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + \frac{1}{n}}\right)^n

Taking the limit as n goes to infinity... this might look like a familiar limit. Then recall that the Radius of convergence is the reciprocal of this limit.
 
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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