# Evaluate $\lim_{n \to \infty}\Bigl[\bigl(\prod_{k=1}^{n}{\frac{2k}{2k-1}}\bigr)\bigl(\int_{-1}^{\infty}\frac{(\cos{x})^{2n}}{2^x}\,dx\bigr)\Bigr]$

The following calculus problem is from the Stanford Math Tournament, 2013 #10.

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

I've had immense difficulty with this problem, and I wasn't really able to understand the solutions to this problem provided by Stanford. Hopefully you guys can help me out. Thanks!

Edit: Okay, to the person who voted my question to be closed because it was deemed off-topic and did not provide enough context: there's not really much I can say about my own approach to this question since it is beyond my level. I'm pretty much clueless about where to even begin, so sorry about that.

• Huh, that's degenerative. Thanks for the heads up, I'll make an edit. – A is for Ambition Jan 2 '15 at 5:37
• The integral is converging to zero, so in order for this question to be non trivial, the product must be going to infinity. – Mark Jan 2 '15 at 5:40
• @Mark: Is there anything we can do with that information? – A is for Ambition Jan 2 '15 at 5:41
• It seems we would like to pull the product inside the integral so that the limit can stabilize. – Mark Jan 2 '15 at 5:44
• Do you think you can show your work to elaborate on this idea? Sorry for my lack of understanding.. – A is for Ambition Jan 2 '15 at 5:47

The product is equal to

$$\frac{2^{2 n}}{\displaystyle \binom{2 n}{n}}$$

which behaves as $\sqrt{\pi n}$ as $n \to \infty$. As for the integral, rewrite as

$$\int_{-1}^0 dx \, 2^{-x} \cos^{2 n}{x} + \sum_{k=0}^{\infty} 2^{-\pi k} \int_0^{\pi} dx \, 2^{-x} \cos^{2 n}{x}$$

Rewrite in a form useful for Watson's Lemma:

$$\int_{-1}^0 dx \, 2^{-x} e^{2 n \log{(\cos{x})}} + \frac1{1-2^{-\pi}}\left [ \int_0^{\pi/2} dx \, 2^{-x} e^{2 n \log{(\cos{x})}}+\int_{\pi/2}^{\pi} dx \, 2^{-x} e^{2 n \log{(-\cos{x})}}\right ]$$

or

$$\int_{-1}^0 dx \, 2^{-x} e^{2 n \log{(\cos{x})}} + \frac1{1-2^{-\pi}}\left [ \int_0^{\pi/2} dx \, 2^{-x} e^{2 n \log{(\cos{x})}}+\int_{0}^{\pi/2} dx \, 2^{x} e^{2 n \log{(\cos{x})}}\right ]$$

The maximum of the argument of the exponential in each integral occurs at $x=0$. About $x=0$, $\log{(\cos{x})} \sim -x^2/2$. By Watson's Lemma, we may extend the interval of integration in each integral out to infinity with exponentially small error as $n \to \infty$. Thus, each integral behaves as $\frac12 \sqrt{\pi/n}$ in this limit. Note, however, that there is no contribution from the integral over $[-1,0]$ because its contribution is subsumed in the first integral over $[0,\pi/2]$ because $1 \lt \pi/2$. Thus, the sought-after limit is

$$\sqrt{\pi n} \frac{1}{1-2^{-\pi}} \sqrt{\frac{\pi}{n}} = \frac{\pi}{1-2^{-\pi}}$$

Likely closer to the problem-poser's intent is the observation that

$$\prod_{k=1}^n \frac{2 k}{2 k-1} = \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}}$$

and, as $n \to \infty$,

$$\int_0^{\pi} dx \, 2^{-x} \cos^{2 n}{x} \sim \int_0^{\pi} dx \, \cos^{2 n}{x} = \frac{\pi }{2^{2 n}} \binom{2 n}{n}$$

The result follows.

• Hi Ron. According to Stanford's provided solutions, the answer is supposed to be $$\pi\frac{2^\pi}{2^\pi - 1}$$ – A is for Ambition Jan 2 '15 at 6:06
• @user155812: I realized my mistake before you posted your comment. I hope my analysis is clear enough. – Ron Gordon Jan 2 '15 at 6:21
• Thank you, Ron. Your solution is very much comprehensive and appreciated. – A is for Ambition Jan 2 '15 at 6:21
• Also, do you think you could explain how/why you are able to rewrite the integral partially using the sigma summation? – A is for Ambition Jan 2 '15 at 6:27
• @user155812: The period of $\cos^2{x}$ is $\pi$, but then I wanted to further rewrite the integral to account for the different behaviors of the cosine over the first and second halves of the interval $[0,\pi]$. – Ron Gordon Jan 2 '15 at 6:33

I will use $I$ as the indicator function.

When $\lim_{n\rightarrow \infty}\cos(x)^{2n} = I[x = \pi k]_{k \in \mathbb{N}}$ so when $n$ is large, the only points that really matter are multiples of $\pi$. For this reason, write $\cos(x) = \cos(x)(I_{S_\epsilon} + I_{S_\epsilon ^c})(x)$ where $S_\epsilon$ is the set of all real numbers within $\epsilon$ distance of a multiple of $\pi$ which are greater than $-1$.

Note $\cos(x)^{2n} = \cos^{2n}(x)I_{S_\epsilon}(x) + \cos^{2n}(x)I_{S_\epsilon ^c}(x)$

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

$$\lim_{n \to \infty}\left[\left(\prod_{k=1}^{n}{\frac{2k}{2k-1}}\right)\left(\int_{I_{S_\epsilon}}\frac{ \cos^{2n}(x)}{2^x}\,dx\right)\right] + \lim_{n \to \infty}\left[\left(\prod_{k=1}^{n}{\frac{2k}{2k-1}}\right)\left(\int_{I_{S_\epsilon ^c}}\frac{ \cos^{2n}(x)}{2^x}\,dx\right)\right]$$

Let's call $p = \cos(\epsilon)$ so that $p \lt 1$. Then $\cos^{2n}(x)I_{S_\epsilon ^c}(x) < p^{2n}.$ Now since the product is bounded by $K\sqrt{n}$ we may compute the second term as

$$\lim_{n \to \infty}\left[\left(\prod_{k=1}^{n}{\frac{2k}{2k-1}}\right)\left(\int_{-1}^{\infty}\frac{ \cos^{2n}(x)I_{S_\epsilon ^c}(x)}{2^x}\,dx\right)\right] \le \lim_{n \to \infty}\int_{-1}^{\infty}\frac{ K p^{2n}\sqrt{n}}{2^x}\,dx = 0$$

where the right hand side $=0$ is obtained by monotone convergence theorem.

Now we estimate the first term $\int_{I_{S_\epsilon}}\frac{( \prod_{k=1}^{n}{\frac{2k}{2k-1}})\cos^{2n}(x)}{2^x}dx$ as $n$ grows large. Now I will be a little sloppy with estimates to motivate the solution. Note the product becomes close to $\sqrt{\pi n}$ as Ron Gordon's solution states. The integral will be close to

$$\int_{I_{S_\epsilon}}\frac{\sqrt{\pi n}\cos^{2n}(x)}{2^x}\,dx$$

$$\sum_{k \ge 0} \int_{\pi k - \epsilon} ^{\pi k + \epsilon}\frac{\sqrt{\pi n}\cos^{2n}(x)}{2^x}\,dx$$

Now concentrate on one of the terms in the sum

$$\int_{\pi k - \epsilon} ^{\pi k + \epsilon}\frac{\sqrt{\pi n}\cos^{2n}(x)}{2^x}\,dx$$

$$\frac{1}{2^{\pi k}}\int_{- \epsilon} ^{ \epsilon}\frac{\sqrt{\pi n}\cos^{2n}(x)}{2^{x}}\,dx$$

Substituting $x = \arcsin(u)$

$$\frac{1}{2^{\pi k}} \int_{-\sin(\epsilon)}^{\sin(\epsilon)}\frac{\sqrt{\pi n}(1 - {u}^2)^{n - \frac{1}{2}}}{2^{\arcsin { u}}}\,du$$

Further substituting $\frac{1}{\sqrt{\pi n}} v = u$

$$\frac{1}{2^{\pi k}} \int_{- \sqrt{\pi n} \sin(\epsilon)}^{\sqrt{\pi n} \sin(\epsilon)}\frac{(1 - \frac{v^2}{\pi n})^{n - \frac{1}{2}}}{2^{\arcsin \frac{1}{\sqrt{\pi n}} v}}\,dv$$

By noting that for some $C \gt 0$, $|\frac{(1 - \frac{v^2}{\pi n})^{- \frac{1}{2}}}{2^{\arcsin \frac{1}{\sqrt{\pi n}} v}} - 1| \lt C \epsilon$ uniformly inside the domain of integration, so we can choose $\epsilon$ small enough so that the term in question is arbitrarily close to 1. Hence we have

$$\approx \frac{1}{2^{\pi k}} \int_{- \sqrt{\pi n} \sin(\epsilon)}^{\sqrt{\pi n} \sin(\epsilon)}(1 - \frac{v^2}{\pi n})^{n}\,dv$$

Write $(1 - \frac{v^2}{\pi n})^{n}$ as $e^{n \log(1 - \frac{v^2}{\pi n})}$. Expanding $\log(1 + x) = x - x^2 g(x)$ where $g(x)$ is analytic around $0$ and $g(0) = 1/2$ We get $e^{n (- \frac{v^2}{\pi n} - \frac{v^4}{\pi^2 n^2} g(\frac{v^2}{\pi n}))} = e^{- \frac{v^2}{\pi}}e^{-\frac{v^4}{\pi^2 n} g(\frac{v^2}{\pi n})}$ Inside the domain of integration $|g(\frac{v^2}{\pi n}) - 1/2| \lt C\epsilon$ and $|\frac{v^4}{\pi n^2}| \le \pi^2 n \sin(\epsilon)^4$ If we allow $\epsilon$ to be a function of $n$ and write $\epsilon_n = \sqrt{\frac{\log n}{n}}$, this will bound $e^{-\frac{v^4}{\pi^2 n}}$ close to 1 inside the region of integration.

$$\approx \frac{1}{2^{\pi k}} \int_{- \sqrt{\pi n}\sqrt{\frac{\log n}{n}}}^{\sqrt{\pi n} \sqrt{\frac{\log n}{n}}}e^{-v^2 / \pi}\,dv$$

$$\approx \frac{1}{2^{\pi k}} \int_{- \infty}^{\infty}e^{-v^2 / \pi}\,dv$$

Now Recognizing the gaussian integral, $$\approx 2^{-\pi k} \pi$$

So taking the sum we get $$=\pi \sum_{k \ge 0} (2^{-\pi})^k$$ $$=\pi \frac{1}{1-2^{-\pi}}$$

Now we must go back to check that our definition of $\epsilon_n$ did not break the first computation of $\lim_{n \to \infty}\int_{-1}^{\infty}\frac{ K p^{2n}\sqrt{n}}{2^x}\,dx$ since $p = \cos(\epsilon)$ is now a function of $n$. But we get $K p_n ^{2n} \sqrt{n} = K ({\log n / n})^n \sqrt{n}$ so we can still use monotone convergence theorem.

There is also the technicality of taking the limit inside the infinite sum, but we can look back on our estimates and see that they were all uniform in $k$ since we did not appeal to the value of $k$ at all.