# Convergence of a Trigonometric Series

After working with a Fourier series for a while, I realized that it would be of great help to me if I could prove that the following limit is zero: $$\lim_{N\to\infty}\frac{1}{N}\sum_{m=1}^\infty\frac{(-1)^m\cos(m(N+1))\sin(mN)}{(4m^2-1)\sin(m)}$$

I tried to do several things and I can show that with a trivial upper bound that if $\sum_{m=1}^\infty\frac{1}{m^2|\sin(m)|}$ converges then the limit will clearly be zero, but this only seems to complicate things, as I think I would have to introduce notions of Diophantine approximation to show that the denominator doesn't get too crazy.

• The numerator is missing a ")". If it's supposed to be $\cos(m(N+1)) \sin(mN)$ then it converges because $|\sin(mN) / \sin(m)| \leq N \ll_N 1$ and the other factors are $O(1/m^2)$. – Noam D. Elkies Jan 27 '14 at 0:07
• Cross post: math.stackexchange.com/questions/652609/… – rghthndsd Jan 27 '14 at 0:09
• @rghthndsd Those two questions don't look the same to me. – David E Speyer Jan 27 '14 at 0:17
• @NoamD.Elkies Let $N$ get arbitrarily large and it is no longer bounded. – Ethan Jan 27 '14 at 0:19
• Is your question whether the sum converges (in which case Noam's answer is right), or whether it converges to a value bounded independently of $N$? – David E Speyer Jan 27 '14 at 0:26

## 1 Answer

As pointed out earlier, $$a_{N,m}:=\frac{(-1)^m\cos(m(N+1))\sin(mN)}{N(4m^2-1)\sin(m)}$$ satisfies $$\left|a_{N,m}\right|\leq \frac{1}{(4m^2-1)}, \mbox{ and }\sum_{n=1}^\infty \frac{1}{(4m^2-1)} < \infty.$$ On the other hand, for every $m$, there holds $$\lim_{N\to\infty} a_{N,m} = 0.$$ Thus the Dominated Convergence Theorem does it for you, $$\lim_{N\to\infty} \sum_{m} a_{N,m} =0.$$ (If you have never seen this Theorem, it says, for series: Given ${a_{N,m}}$,, if

• $|a_{N,m}| < b_m$ for all $N$ and $\sum_{m} b_m$ converges
• for $m$ fixed, $a_{N,m}\to c_m$

then

• $\sum_{m} |a_{N,m} -c_m| \to 0$
• $\sum_m c_m$ converges

and consequently $\lim_{N\to \infty} \sum_m a_{N,m} = \sum_m c_m$).