I’d like to show for some $c,N>0$ and all positive integer $n>N$ that
$$\int^\infty_{-\infty} \frac{\sin(x/n)}{x} \prod_{i=1}^n \cos\left(\frac xi\right)dx>\frac cn$$
If I just look at the integral near $0$, I can get that for $x$ much less than $1$, the cosine terms are positive and approximately quadratic and the sin term is odd and approximately linear, so the integrand is positive and around $(1-\pi^2 x^2/6)/n$ which when integrating from $-1$ to $1$ gives a positive value proportionate to $1/n$, but I’m having trouble bounding the potentially large oscillations of the tails.
I have noticed that $x$ and $2\pi k n! -x$ give the same value for the cosine product and opposite signs for the sine, so this could lead to a significant amount of cancellation, but making this rigorous seems tricky.
This problem came up while I was looking at the characteristic function of the partial sums of a random variant of the harmonic series while digging into this problem: How often do we expect a random walk with decreasing step size to cross $0$?
Best Answer
Equivalently, this means $n$ times the proportion of $(e_1,…,e_n)\in \{−1,1\}^n$ such that $\sum_{k=1}^n e_k/k$ belongs to $[−1/n,1/n]$ is bounded below.
That is not too hard. We'll just use the cheap observation that if $2^m\le n<2^{m+1}$, then $\sum_{j=0}^m \frac{e_{2^j}}{2^j}$ has uniform distribution on the arithmetic progression of $2^{m+1}$ terms with step $2^{1-m}$ running essentially from $-2$ to $2$.
Now, if the requested interval were $[-2/n,2/n]$, that would be the end of the story because the expectation of the square of the rest of the series is less than $\sum_{k\ge 3}\frac 1{k^2}\le \sum_{k\ge 3}\frac 1{(k-1)k}=\frac 12$, so with probability $\ge \frac 12$, the rest adds up to a number in $[-1,1]$ and the shift of our arithmetic progression by that number would have at least one term in the desired interval (because $2^{1-m}<\frac 4n$), yielding the lower bound $2^{-m-2}\ge\frac 1{4n}$.
In our case the interval is a bit shorter, so for large $m$ and $n$ we will also consider $e_p,e_q$ with $\frac{2^{m-1}}p\approx \frac 54, \frac{2^{m-1}}q\approx \frac 98$. Then considering the contribution of $\frac{e_p}{p}+\frac{e_q}q$ as well, we see that we have a small perturbation of an arithmetic progression with step $2^{-m-1}<\frac 1n$, so we are still fine but with the bound $\frac 1{16n}$.
That takes care of large $n$, so all that remains is to show that the probability is never $0$. That can be achieved by the greedy choice of signs going in the direction of $0$ every time. The trivial induction shows that we'll be under or at $1/j$ in absolute value after adding $\pm 1/j$.
It is also worth mentioning that the integral is not this probability, but this probability minus one half of the probability that the sum is exactly $\pm 1/n$ (the Fourier integral for a function with nice jump discontinuity converges to the value in the middle of the jump). That doesn't change much though: the final bound merely drops at most twice more.