# Asymptotic related to the infinite product of sine

The amount is somewhat complicated ($x$ is a constant):

$$S_n=\sum_{k=1}^n\ln\left(1-\frac{\sin^2\big(x/(2n+1)\big)}{\sin^2\big(k\pi/(2n+1)\big)}\right)\tag{*}$$

I want to enrich my handy powerful tools to discover the asymptotic behavior of such a summation directly as sharp as possible.

Such an amount could be reduced in such a way:

Since $(\cos x+i\sin x)^n=\cos nx+i\sin nx$, it's not hard to show that

$$\sin(2n+1)\theta=\sin\theta\cdot P_n(\sin^2\theta)$$

where $P_n$ is a polynomial depend on $n$.

From the zeros of $P_n(x)$, we could easily obtain that

$$P_n(x)=(2n+1)\prod_{k=1}^n\left(1-\frac x{\sin^2\big(k\pi/(2n+1)\big)}\right)$$

Therefore

$$S_n=\ln\sin x-\ln(2n+1)\sin\frac x{2n+1}\tag{1}$$

However, I want to obtain some techniques to estimate the summation directly. The methods and the disciplines are much more important than the result. As a result, one can compare with identity (1) to obtain some useful results. That's common in Eulerian mathematics.

If $k$ is small enough, i.e $k=o(n)$, the summand

$$a_{k}(n)=\ln\left(1-\frac{\sin^2\big(x/(2n+1)\big)}{\sin^2\big(k\pi/(2n+1)\big)}\right)$$

is approximately

$$\ln\left(1-\frac{x^2}{k^2\pi^2}\right)$$

From some mechanical computation (Taylor expansion, etc), we could obtain a stronger result (if there's no calculation mistakes):

$$a_k(n)=\ln\left(1-\frac{x^2}{k^2\pi^2}\right)+\frac{x^2}{3(2n+1)^2}+O(k^2n^{-4})$$

We could split the sumation $S_n=\sum_{k=1}^n=\sum_{0<k<n^{1-\epsilon}}+\sum_{n^{1-\epsilon}\le k\le n}$, where $\epsilon>0$ is a small constant, then the major distribution $\sum_{0<k<n^{1-\epsilon}}a_k(n)$ could be easily controlled, estimated as good as we want; however, the tail $\sum_{k\ge n^{1-\epsilon}}a_k(n)$ is a bit messy. It's not very small. I can only show that $\sum_{k\ge n^{1-\epsilon}}a_k(n)=O(n^{-1+2\epsilon})$, therefore the best result I could get is:

$$S_n=\sum_{k=1}^\infty\ln\left(1-\frac{x^2}{k^2\pi^2}\right)+O(n^{-1+2\epsilon})$$

which is strong enough to prove the infinite product of $\sin$ function.

However, I want to look into the $O$ notation to obtain sharper and sharper results. Can we do that directly through the definition (*)?

Thanks!

EDIT: $x$ might be a real constant (for my ignorance), but one could assume that it's a complex constant. The question of the real version is just the special case of the complex version, and there's no doubt that I'm pleased if one can deal with the complex version well.

Remark: The case is somewhat like the case when Euler-Maclaurin formula is applicable. In that case, there's no tail after all! Here, the tail is small, but it's not small enough, say, exponentially small. I guess the key is to use an integral to estimate the error.

• What kind of results are you hoping for by comparing the asymptotic with the closed form $(1)$? (Also, what is $x$ in the closed form?) – Antonio Vargas May 2 '13 at 16:02
• @AntonioVargas $x$ is a constant, as I've said. What I really need, is methods, discipline to deal with such a summand. Comparing is just yielding by-products. – Yai0Phah May 3 '13 at 8:20
• @AntonioVargas I think I've got the key to attack the problem. Substract $S_n$ with $\sum_{k=1}^n\ln(1-x^2/k^2\pi^2)$, and estimate the summation through Euler-Maclaurin formula. You can plot a graph for the summation, and find that the distribution is very flat, which case is suitable for Euler-Maclaurin! Detailed calculation is not performed since the calculation is tedious. – Yai0Phah May 10 '13 at 16:12