Proving $\displaystyle\lim_{m\rightarrow \infty}\sqrt {2m+1}\int_{0}^{1}\frac{t^{(2m+1)n}}{\sqrt {1-t^{2n}}}\,dt\ $ exists and is positive I’m new here, sorry if my question isn’t formatted correctly. I’m stuck on exercise 1.2.1 in this: http://users.mai.liu.se/vlatk48/teaching/lect2-agm.pdf (page 16), which asks the reader to prove the convergence of the following limits:
$$\lim_{m\rightarrow \infty}\frac{1}{\sqrt {2m+1}}\prod_{j=1}^{m}\frac{2jn+1}{(2j-1)n+1}$$
and
$$\lim_{m\rightarrow \infty}\sqrt {2m+1}\int_{0}^{1}\frac{t^{(2m+1)n}}{\sqrt {1-t^{2n}}}\,dt$$
exist and are greater than 0. 
I can do the product; it’s just expressing $\frac{1}{\sqrt {2m+1}}$ as a product from 1 to m, squaring the whole expression, and performing the sum test, but I can’t figure out the integral. The best I’ve gotten is that I can reduce it to a special case.
First, note that since the integral runs from 0 to 1,  $\sqrt {1-t^{2n}}$ is larger for larger n, while $t^{(2m+1)n}$ is smaller for larger n, so the expression converges for all n if and only if it converges for 1.
Then, make the following change of variables 
$$u=t^n$$
$$dt=\frac{du}{nt^{n-1}}$$
The division by $t^{n-1}$ only makes the integrand larger, since t is between 0 and 1, and n is a constant with respect to the limit, so the limit is greater than 0 for all n iff it is greater than 0 for n=1. 
But despite thinking about it all day, I can’t even begin to figure out how to crack the special case. Can someone point me in the right direction?
 A: I believe I've figured it out. Since I've already shown how this reduces to a special case, let's start by assuming n is 1, giving us
$$\lim_{m\rightarrow \infty}\sqrt {2m+1}\int_{0}^{1}\frac{t^{2m+1}dt}{\sqrt {1-t^{2}}}$$
Make the change of variables 
$$t=sin(\theta)$$
$$dt=cos(\theta)d\theta $$
which leaves 
$$\lim_{m\rightarrow \infty}\sqrt {2m+1}\int_{0}^{\frac{\pi}{2}} sin(\theta)^{2m+1}d\theta=\lim_{m\rightarrow \infty}\sqrt {2m+1}\frac{1}{2}B(m+1,\frac{1}{2})$$
Now, writing the beta function in terms of the gamma function, we have
$$\frac{1}{2}B(m+1,{\frac{1}{2}})=\frac{\Gamma(m+1)\Gamma(\frac{1}{2})}{2\Gamma(m+\frac{3}{2})}=\frac{\Gamma(m+1)\sqrt{\pi}}{2\Gamma(m+\frac{3}{2})}$$
Using Stirling's approximation for both of the gamma functions, this asymptotically approaches 
$$\frac{\sqrt{\pi}}{2\sqrt{m+1}}$$
as m tends to infinity.Plugging this back into our original equation gives a nonzero limit of $\frac{\sqrt{2\pi}}{2}$ for the case of n=1, which proves that some limit exists for the other cases. I'd be happy for someone to check my work, but I think this is it. Thanks to Martin Argerami for pointing me towards Wolfram Alpha.
Edit: Corrected errors.
A: Let $v=t^{2n}$. Then $dv=2nt^{2n-1}\,dt$. By definition of the Beta Function, and expressing the Beta function interms of the Gamma function, your integral is 
$$
\int_{0}^{1}\frac{t^{(2m+1)n}}{\sqrt {1-t^{2n}}}\,dt=\frac1{2n}\int_0^1v^{m+\frac12}(1-v)^{-1/2}\,dv=\frac1{2n}\,B(m+\tfrac32,\tfrac12)=\frac{\Gamma(m+\tfrac32)\Gamma(\tfrac12)}{2n\Gamma(m+2)}.
$$
Thus, using Stirling's Approximation, we get for big $m$
\begin{align}
\sqrt{2m+1}\int_{0}^{1}\frac{t^{(2m+1)n}}{\sqrt {1-t^{2n}}}\,dt
&= \frac{\sqrt\pi}{2n}\,\frac{\sqrt{\frac{2\pi}{m+3/2}}\left(\frac{m+3/2}e\right)^{m+3/2}}{\sqrt{\frac{2\pi}{m+2}}\left(\frac{m+2}e\right)^{m+2}}\,\sqrt{2m+1}\\ \ \\
&= \frac{\sqrt{\pi e}}{2n}\,\frac{\sqrt{\frac{2\pi}{m+3/2}}\,\left({m+3/2}\right)^{m+3/2}}{\sqrt{\frac{2\pi}{m+2}}\,\left( {m+2}\right)^{m+3/2}}\,\frac{\sqrt{2m+1}}{\sqrt{m+2}}\\ \ \\
&\xrightarrow[m\to\infty]{}\frac{\sqrt{\pi e}}{2n}\,1\,\frac1{\sqrt e}\,\sqrt2
=\sqrt{\frac{\pi}{2n^2}}
\end{align}
I find all this fairly sophisticated for a casual exercise, but I didn't check your source carefully for context. 

Comment 1:
If your sources for the beta function differ from the Wikipedia article (since you used the sine), you can push your formula to arbitrary positive $n$: you simply do the substitution $t^{n}=\sin\theta$. Then $nt^{n-1}\,dt=\cos\theta\,d\theta$. So
\begin{align}
\int_{0}^{1}\frac{t^{(2m+1)n}}{\sqrt {1-t^{2n}}}\,dt
&=\int_{0}^{\pi/2}\frac{t^{(2m+1)n}}{nt^{n-1}}\,d\theta
=\frac1n\,\int_{0}^{\pi/2}{\sin^{2mn+1}}\theta\,d\theta\\ \ \\
\end{align}

Comment 2: a manipulation with partial fractions that removes the singularity at 1. 
\begin{align}
\int_{0}^{1}\frac{t^{(2m+1)n}dt}{\sqrt {1-t^{2n}}}
&=\int_{0}^{1}\frac{t^{2mn-n+1}\,t^{2n-1}dt}{\sqrt {1-t^{2n}}}\\ \ \\
&=-\left.\vphantom{\int}\frac {t^{2mn-n+1}\sqrt {1-t^{2n }}}{n}\right|_0^1
+\tfrac {2mn-n+1}{n}\int_0^1t^{2mn-n}\sqrt {1-t^{2n}}\,dt\\ \ \\
&=\tfrac {2mn-n+1}{n}\int_0^1t^{2mn-n}\sqrt {1-t^{2n}}\,dt.
\end{align}
