Find the radius of convergence

JulioMarcos
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Homework Statement
Find the radius of convergence of the Series:
\sum_{i=1}^{\infty}\frac{(2n)!x^n}{(n!)^2}

The attempt at a solution
I used the Ratio Test but I always get L = |\frac{2x}{n+1}|

The answer is 1/4. I think I am mistaking with factorial.
 
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Solved. The problem was tha
(2(n+1))! = (2n + 2)(2n + 1)(2n)!
and I was doing (2(n+1))! = (2n + 2)(2n)!
because i thought you should distribute the 2
 
Ah it is always good to find your own mistake. Off the bat that was my guess... Sadly I got in the thread too late.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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