Compute $\lim\limits_{n\to\infty} \int_{0}^{2\pi} \cos x \cos 2x\cdots \cos nx \space{dx}$ 
Compute the following limit:
  $$\lim_{n \to \infty}\int_{0}^{2\pi}\cos\left(x\right)\cos\left(2x\right)\ldots
\cos\left(nx\right)\,{\rm d}x$$

Today I was working on a W. L. Putnam competition's problem containing this 
integral and wondered how I may compute its value when $n$ goes to $\infty$. So far I've found no answer.
Could you give me some suggestions about the way I should go?   
 A: We have
$$\begin{array}{c l}\int_0^{2\pi}\cos x\cdots\cos nx dx= & \int_0^{2\pi}\prod_{k=1}^n\frac{e^{ikx}+e^{-ikx}}{2}dx \\ & =\frac{1}{2^n}\int_0^{2\pi}e^{-n(n+1)/2\,x}\prod_{k=1}^n(1+e^{i2kx})dx \\ & = \frac{1}{2^n}\sum_{K\subseteq [n]}\int_0^{2\pi}\exp\left(-\frac{n(n+1)}{2}i+2i\sum_{k\in K}k\right)dx \\ & =\frac{2\pi|L_n|}{2^n}. \end{array}$$
where $L_n$ is the set of subsets $K$ of $[n]=\{1,\cdots,n\}$ such that $$2\sum_{k\in K}k=\frac{n(n+1)}{2}.$$
(Note this is zero when $n\ne 0,-1$ mod $4$.) Perhaps one can show $|L_n|=\Omega(2^n)$ with a combinatorial argument? Since it's $0$ infinitely often it would also suffice to show the limit exists.
A: You can prove instead that 
$$\lim_{n\to\infty} \int_{0}^{2\pi} \left|  \cos x \cos 2x\cdots \cos nx \space{dx} \right| = 0 \,.$$
Let $\epsilon >0$. Then for all $n$ we have
$$\int_{0}^{2\pi} \left|  \cos x \cos 2x\cdots \cos nx \right| \space{dx} = \int_{0}^{\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx \right| \space{dx} $$
$$+ \int_{\frac{\epsilon}5}^{\pi-\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx} + \int_{\pi-\frac{\epsilon}5}^{\pi+\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx}$$
$$ + \int_{\pi+\frac{\epsilon}5}^{2\pi-\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx}+\int_{2 \pi-\frac{\epsilon}3}^{2\pi} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx}$$
$$\leq \frac{4\epsilon}{5}+\int_{\frac{\epsilon}5}^{\pi-\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx}+  \int_{\pi+\frac{\epsilon}5}^{2\pi-\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx}$$
I split it this way to avoid the points $x$ where $\cos(kx) \in \{ \pm 1 \}$.
Now, we claim that for all $x \in [\frac{\epsilon}{5},  \pi-\frac{\epsilon}{5}] \cup  [\pi+\frac{\epsilon}{5},  2\pi-\frac{\epsilon}{5}]$ we have
$$\lim_n \left|  \cos x \cos 2x\cdots   \right| =0$$
This is easy to prove, by splitting it in two cases: x rational and irrational with respect to $\pi$. In the first case, it follows from the Dirichclet Principle that there exists a $k$ so that for all $n$ one of $|\cos nx| , |\cos (n+1)x|, ..., |\cos (n+k)x|$ is less than $\frac{1}{2}$, while the second case is easy to threat by writing $x= \frac{k}m \pi$ with $gcd(m,k)=1$ and $m\neq 1$.
To complete the proof, all these functions are uniformly bounded by $1$ and they converge pointwise to $0$, thus by Lebesgue dominant convergence theorem we have
$$\int_{\frac{\epsilon}5}^{\pi-\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx}+  \int_{\pi+\frac{\epsilon}5}^{2\pi-\frac{\epsilon}5} \left|  \cos x \cos 2x\cdots \cos nx  \right| \space{dx} \to 0$$
P.S. This proof is probably an overkill...
A: Let $X_j$ be independent coin flips ($0$ or $1$ with equal probabilities) and 
$S_n = \sum_{j=1}^n j X_j$.  From Thomas Andrews's answer, we see that 
$\int_0^{2\pi} \cos(x) \cos(2x) \ldots \cos(nx)\ dx = 2 \pi P(S_n = n(n+1)/4)$.
We have $E[S_n] = n(n+1)/4$ and $\text{Var}(S_n) = \dfrac{1}{4} \sum_{j=1}^n j^2 = \dfrac{n(n+1)(2n+1)}{24}$. Moreover, Lindeberg's version of the Central Limit Theorem (which does not require the random variables to be identically distributed) applies.  We
conclude that for any $\alpha > 0$, $P(|S_n - n(n+1)/4| < \alpha \sqrt{n(n+1)(2n+1)/24}) \to \Phi(\alpha) - \Phi(-\alpha)$ as $n \to \infty$, where $\Phi$ is the CDF of the standard normal distribution.  Since $\Phi(\alpha) - \Phi(-\alpha) \to 0$ as $\alpha \to 0$,
we conclude that $P(S_n = n(n+1)/4) \to 0$.
A: Let $u = e^{ix}$.  Then $\cos x \cos 2x ... \cos nx = \frac{1}{2^n}\prod_{k=1}^n (u^k + u^{-k})$.
Write $$\prod_{k=1}^n (z^k + z^{-k}) = \sum_{j=-N}^N a_j z^j$$
where $N=\frac{n(n+1)}2$. 
Then the integral above is just $\frac{2\pi}{2^n}a_0$.
Combinatorially, we can see that $a_0$ is the number of ways of picking a subset of $\{1,2,...,n\}$ which adds up to $\frac{n(n+1)}{4}$.
Now, the set of subsets of $\{1,...,n\}$ that add up to a particular value is an antichain, so we have a bound on the size of $a_0$, namely, $$0\leq a_0 \leq \binom {n}{\lfloor n/2\rfloor}$$.
And we know that $$\lim_{n\to\infty}\frac{1}{2^n}\binom {n}{\lfloor n/2\rfloor} =0$$
So we know your limit is zero.
(The maximum size of an antichain in the Boolean poset is Sperner's theorem.)
Note that when $n\equiv 1,2\pmod 4$, $\frac{n(n+1)}4$ is not an integer, so in those cases, $a_0=0$.  Anybody have an "interesting" lower bound on $a_0$ for the other cases?
[In comments below, Robert Israel shows that when $n\equiv 0,3\pmod 4$, $a_0\geq 2^{\lfloor (n+1)/4\rfloor}$. I suspect that there is a polynomial, $q$, such that $a_0\geq \frac{2^n}{q(n)}$.]
This proof generalizes.  Given any sequence of positive integers, $\{b_1,...,b_n\}$, we have the same result:
$$\int_{0}^{2\pi} \cos b_1 x \cos b_2 x ... \cos b_n x dx= \frac{2\pi}{2^n}A$$
where $A$ is the count of an antichain, so $0\leq A\leq \binom {n}{\lfloor n/2\rfloor}$
If $b_i=1$ for all $i$, then we get exact equality when $n$ is even, which is to say that:
$$\int_{0}^{2\pi} \cos^{2m} x dx = \frac{2\pi}{2^{2m}} \binom {2m}{m}$$
A: Another analysis proof:
It suffices to show that
$$\lim_{n \rightarrow \infty} \int_0^{2\pi}|\cos x \cos 2x\cdots \cos nx|\,dx = 0$$
Define $f_n(x) = |\cos x \cos 2x\cdots \cos nx|$. Then since each $|\cos kx | \leq 1$, $f_n(x) \geq f_{n+1}(x)$ for all $n$ and $x$. Since $|f_n(x)| \leq 1$ for all $n$ and $x$, the Lebesgue dominated convergence theorem says
$$\lim_{n \rightarrow \infty} \int_0^{2\pi} |\cos x \cos 2x\cdots \cos nx|\,dx = 
\int_0^{2\pi} \lim_{n \rightarrow \infty}|\cos x \cos 2x\cdots \cos nx|\,dx$$
So it suffices to show that the limit on the inside is zero for almost every $x$. But whenever $x$ is not a rational multiple of $\pi$, $\{nx: n = 1,2,..\}$ is dense in $[0,2\pi]$. Which means given such $x$, there is a sequence $n_1, n_2,...$ such that
$$\lim_{k \rightarrow \infty} n_k x = {\pi \over 2}$$
 So 
$$\lim_{k \rightarrow \infty} \cos(n_k x) = \cos({\pi \over 2}) = 0$$
 Since $\lim_{n \rightarrow \infty}|\cos x \cos 2x\cdots \cos nx|$ is at most any given $|\cos (n_k x)|$, we conclude that  $\lim_{n \rightarrow \infty}|\cos x \cos 2x\cdots \cos nx| = 0$ for all $x$ not a rational multiple of $\pi$. Since the multiples of $\pi$ are a set of measure zero, we are done.
A: Choose $0<\epsilon<\pi$. Set $f_n(x)=|\cos(x)\cos(2x)\cdots\cos(nx)|$. Clearly the sequence of (continuous) functions $(f_n)$ converges pointwise to $0$ on $I_{\epsilon}=[\epsilon,2\pi-\epsilon]$, and $0\leq f_{n+1}\leq f_n\leq 1$. It follows from Dini's lemma that $f_n$ converges uniformly to $0$ on $I_{\epsilon}$, so there exists an integer $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, $0\leq f_n\leq\epsilon$. Thus, for all $n\geq N_{\epsilon}$,
$$\begin{align*}\left|\int_0^{2\pi}f_n \right|&\leq \left|\int_0^{\epsilon}1 \right|+\left|\int_{\epsilon}^{2\pi-\epsilon}f_n \right|+\left|\int_{2\pi-\epsilon}^{2\pi}1 \right|
\\ &\leq (2+2\pi)\epsilon\end{align*}$$
Hence
$$\int_0^{2\pi}\cos(x)\cos(2x)\cdots\cos(nx)\;dx\xrightarrow{n\to\infty}0\;.$$
