Evaluate $\lim_{n\to\infty} \prod_{k=1}^n \frac{2k}{2k-1}\int_{-1}^{\infty} \frac{{\left(\cos{x}\right)}^{2n}}{2^x} \; dx$ Problem 9 in the JHMT 2013 Calculus Test asks to evaluate
$$\lim_{n\to\infty} \prod_{k=1}^n \frac{2k}{2k-1}\int_{-1}^{\infty} \frac{{\left(\cos{x}\right)}^{2n}}{2^x} \; dx$$
The answer is $\pi\cdot 2^\pi /(2^{\pi}-1)$. How can I show this?  I know that the infinite product diverges and the limit cannot be moved into the integral, but I don't know what to do.  Maybe I can represent the integral as a summation?
 A: I think it's simpler to evaluate the integral like this: $$\ $$ We know that by Wallis formula $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos x)^{2n}=I_n=\frac{2n-1}{2n}I_{n-1}$$ which on recursive application gives us $$I_n=I_0\prod_{k=1}^n\frac{2k-1}{2k}$$ which gives d$$I_n=\pi\prod_{k=1}^n\frac{2k-1}{2k} \space (\text{as}\space I_0=\pi)$$ and as $n\to\infty$ the value of $$\int_{-1}^{\infty}\frac{(\cos x)^{2n}}{2^x}\mathrm{d}x$$ will get concentrated near the values where $\cos x$ becomes $+1$ or $-1$ and that happens at $0,\pi,2\pi,...$
and the area near other parts of the graph will tend to zero . (I understand that this isn't the most rigorous way to put it, but I believe such ideas are based off the Dominated Convergence Theorem, which I am not very familiar with.) However, answers provided by Oliver Diaz and and River Li give a firm proof for this reasoning. Do look through them for thorough assurance of the idea. For $n=10^{9}$the graph is like this(from desmos)So, we can write the integral as $$\sum_{k=0}^{\infty}\frac{I_n}{2^{k\pi}}$$ and the total value as $n\to \infty$ becomes equal to $$\prod_{k=1}^n\frac{2k}{2k-1}\int_{-1}^{\infty}\frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\to \prod_{k=1}^n\frac{2k}{2k-1}\sum_{k=0}^{\infty}\frac{I_n}{2^{k\pi}}=\frac{\pi}{1-2^{-\pi}}=\frac{\pi2^{\pi}}{2^{\pi}-1} $$ and this is valid as long as the lower limit of the integral $$\int_{-1}^{\infty}\frac{(\cos x)^{2n}}{2^x}\mathrm{d}x$$ more than -$\pi$ and if it's less than $-\pi$ then the lower limit of the summation will become $k=-1$ instead of $k=0$
A: Firstly split it up into two parts:
$$\prod_{k=1}^n\frac{2k}{2k-1}=\frac{2.4.6.8...2n}{1.3.5.7.(2n-1)}=\frac{2^nn!\times2^{n-1}(n-1)!}{(2n-1)!}=\frac{2^{2n-1}n!(n-1)!}{(2n-1)!}=\frac{2^{2n-1}(n!)^2}{n(2n-1)!}$$
now the integral:
$$I_n=\int_{-1}^\infty\frac{(\cos x)^{2n}}{2^x}dx$$
$$I_n(a)=\int_{-1}^\infty e^{-ax}\cos^{2n}xdx$$
and we know that:
$$\cos^{2n}x=\frac{(e^{ix}+e^{-x})^{2n}}{2^{2n}}$$
and:
$$(e^{ix}+e^{-ix})^{2n}=\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2n-r)ix}e^{-rix}=\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2n-2r)ix}$$
so our integral becomes:
$$I_n(a)=\int_{-1}^\infty\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2n-2r)ix-ax}dx=I_n(a)=\int_{-1}^\infty\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2i(n-r)-a)x}dx$$
assuming we can interchange the integral and summation and allowing $-b=2i(n-r)-a$ we get:
$$I_n(a)=\sum_{r=0}^{2n}{{2n}\choose{r}}\int_{-1}^\infty e^{-bx}dx=\sum_{r=0}^{2n}{{2n}\choose{r}}\left[\frac{-e^{-bx}}{b}\right]_{-1}^\infty=\sum_{r=0}^{2n}{{2n}\choose{r}}\frac{e^b}{b}$$
$$I_n(a)=\sum_{r=0}^{2n}{{2n}\choose{r}}\frac{e^{a-2i(n-r)}}{a-2i(n-r)}$$
If we bring it all together we get:
$$L=\lim_{n\to\infty}\frac{2^{2n-1}(n!)^2}{n(2n-1)!}\sum_{r=0}^{2n}{{2n}\choose{r}}\frac{e^{\ln(2)-2i(n-r)}}{\ln(2)-2i(n-r)}$$
and we know that:
$${2n\choose r}=\frac{(2n)!}{r!(2n-r)!}=\frac{2^nn!}{r!(2n-r)!}$$
so:
$$L=\lim_{n\to\infty}\frac{2^{3n}(n!)^3}{n(2n-1)!}\sum_{r=0}^{2n}\frac{e^{-2i(n-r)}}{\ln(2)-2i(n-r)}\times\frac{1}{r!(2n-r)!}$$
