Radius of Convergence for Moderately Complicated Series

LukeMiller86
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1. The problem statement:

Show that the following series has a radius of convergence equal to exp\left(-p\right)

Homework Equations



For p real:

\Sigma^{n=\infty}_{n=1}\left( \frac{n+p}{n}\right)^{n^{2}} z^{n}

The Attempt at a Solution


\stackrel{lim}{n\rightarrow\infty}\left|a_{n}\right|^{1/n} = \frac{1}{R} = \left(\frac{n+p}{n}\right)^{n}<br /> =exp\left(n\left(ln\left(\frac{n+p}{n}\right)\right)\right)

Apart from playing with the logarithm after that I cannot seem to reach the required answer.
Any help would be greatly appreciated.
 
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What's the limit definition of the exponential function?
 
exp\left(-p\right) = e^{\left(-p)\right}

is that what you meant?
 
Do you know this limit:
\lim_{n\to\infty}\left(1+\frac{p}{n}\right)^n
 
Completely overlooked that! Thanks very much.
 
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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