$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} dx$ 
Evaluate $$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} dx$$

My try: 
$$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} dx=\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {e^{i2nx}(1+e^{-i2x})^{2n}}{2^{2n}e^{x\ln 2}} dx$$
I write this using that $\cos x=\frac {e^{ix}+e^{-ix}}{2}$ and $2^x=e^{x\ln 2}$
We also know that $$\prod_{k=1}^n \frac {2k}{2k-1}=\frac {2^{2n}(n!)^2}{(2n-1)!}$$ Using this along with binomial theorem we get $$\lim_{n\to \infty} \prod_{k=1}^n \left( \frac {2k}{2k-1}\right) \int_{-1}^{\infty} \frac {e^{i2nx}(1+e^{-i2x})^{2n}}{2^{2n}e^{x\ln 2}} dx=\lim_{n\to\infty} \frac {(n!) ^2}{(2n-1)!}\left(\sum_{r=0}^{2n} \binom {2n}{r}\left(\int_{-1}^{\infty} e^{x(2i(n-r)-\ln 2)} dx\right)\right) $$
$$=\lim_{n\to\infty} \frac {(n!) ^2}{(2n-1)!}\left(\sum_{r=0}^{2n} \binom {2n}{r} \left[\frac {e^{x(2i(n-r)-\ln 2}}{ 2i(n-r)-\ln 2)} \right]_{-1}^{\infty}\right)$$
And now I am stuck here. Any suggestions or a different method are openly welcomed. 

We introduce the following result.

Proposition. Let $f$ be a bounded measurable function on $[-\pi/2, \pi/2]$ which is continuous at $0$. Then
$$ \lim_{n\to\infty} \left(\prod_{k=1}^n\frac{2k}{2k-1}\right) \int_{-\pi/2}^{\pi/2}f(x)\cos^{2n}(x)\,dx = \pi f(0). $$

We defer the proof to the end and rejoice its consequence now. We have
\begin{align*}
\int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} \, dx
&= \int_{-1}^{\pi/2} \frac {(\cos x)^{2n}}{2^x} \, dx + \sum_{k=1}^{\infty} \int_{-\pi/2}^{\pi/2} \frac {(\cos x)^{2n}}{2^{x+k\pi}} \, dx \\
&= \int_{-\pi/2}^{\pi/2} \left( \frac{1}{2^x} \mathbf{1}_{[-1,\pi/2]}(x) + \sum_{k=1}^{\infty} \frac {1}{2^{x+k\pi}} \right) (\cos x)^{2n} \, dx.
\end{align*}
So by the above proposition,
\begin{align*}
\left(\prod_{k=1}^n\frac{2k}{2k-1}\right) \int_{-1}^{\infty} \frac {(\cos x)^{2n}}{2^x} \, dx
&\xrightarrow[n\to\infty]{}
\pi \sum_{k=0}^{\infty} \frac {1}{2^{k\pi}}
= \frac{\pi 2^{\pi}}{2^{\pi} - 1}.
\end{align*}

Proof. By the substitution $x = u/\sqrt{n}$, we may write
$$
\int_{-\pi/2}^{\pi/2} f(x)\cos^{2n}(x) \, dx
= \frac{1}{\sqrt{n}} \underbrace{ \int_{-\pi\sqrt{n}/2}^{\pi\sqrt{n}/2} f\left(\frac{u}{\sqrt{n}}\right)  \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) \, du}_{\text{(*)}}.
$$
Now we make several observations.


*

*It is easy to see from Stirling's formula that $\prod_{k=1}^{n} \frac{2k}{2k-1} \sim \sqrt{\pi n}$, see this for instance.

*There exists $c > 0$ for which $\cos x \leq 1 - cx^2$ for all $|x| \leq \pi/2$. For instance, one may utilize the inequality $\sin x \geq \frac{2}{\pi}x$, valid for $0 \leq x \leq \pi/2$, to show that $c = \frac{1}{\pi}$ works. Together with the inequality $1+x \leq e^x$ which holds for all $x \in \mathbb{R}$,
$$ \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) \leq \left(1 - \frac{cu^2}{n} \right)^n \leq e^{-cu^2} $$
for each $u$ satisfying $|u| \leq \pi\sqrt{n}/2$. This shows that the integrand of $\text{(*)}$, extended to all of $\mathbb{R}$ by setting its value to $0$ outside $[-\pi\sqrt{n}/2, \pi\sqrt{n}/2]$, is bounded by an integrable dominating function.

*For each fixed $u$, Taylor's theorem tells that
$$ f\left(\frac{u}{\sqrt{n}}\right)  \cos^{2n}\left(\frac{u}{\sqrt{n}}\right)
= (f(0) + o(1)) \left( 1 - \frac{u^2}{2n} + \mathcal{O}\left(\frac{1}{n^2}\right) \right)^{2n}
\xrightarrow[n\to\infty]{} f(0) e^{-u^2}. $$
Combining altogether, by the dominated convergence theorem,
$$ \int_{-\pi\sqrt{n}/2}^{\pi\sqrt{n}/2} f\left(\frac{u}{\sqrt{n}}\right)  \cos^{2n}\left(\frac{u}{\sqrt{n}}\right) \, du
\xrightarrow[n\to\infty]{} \int_{-\infty}^{\infty} f(0)e^{-u^2} \, du = \sqrt{\pi}f(0). $$
Together with the observation $\frac{1}{\sqrt{n}} \prod_{k=1}^{n} \frac{2k}{2k-1} \to \sqrt{\pi} $, the desired conclusion follows.
