Finding the Largest Value of b for Convergence

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Find the largest value of b that makes the following statement true: "if 0<= a <= b, then the series (from n=1 to infinity) of (((n!)^2a^n)/(2n!)) converges".

I know you have to do the ratio test for this one but I don't know how to do it.
 
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OK let's call the summand s_n. So we have:

s_n=\frac{n!^{{2a}^n}}{(2n)!}

Can you write down s_{n+1}?
 
im sry its (n!)^2(a^n) for the numerator
 
Yes, use the ratio test.
\left[\frac{((n+1)!)^2a^{n+1}}{(2(n+1))!}\right]\left[\frac{(2n)!}{((n!)^2a^n}\right]= \left[\frac{(n+1)!}{n!}\right]^2\left[\frac{a^{n+1}}{a^n}\right]\left[\frac{(2n)!}{(2(n+1))!}]
= \frac{(n+1)(a)}{(2n+2)(2n+1)
What is the limit of that as n goes to infinity? If that limit depends on a, then the series will converge only for values of a that make the limit less than 1.
 
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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