Uniform convergence of the series $\sum_{n=1}^{\infty} \frac{\cos(2nt)}{4 n^2 - 1} $

Question

I am trying to find if this series is uniformly convergent: $$\sum_{n=1}^{\infty} \frac{\cos(2nt)}{4 n^2 - 1} $$ So far I have (using the Weierstrass M-Test): $$| \frac{\cos(2nt)}{4 n^2 - 1}| \le \frac{-1}{4 n^2 -1}$$ Now using the ratio test for: $$a_n = -\frac{1}{4 n^2 -1}$$ $$\frac{|a_{n+1}|}{|a_n|} = \frac{4n^2 -1}{4(n+1)^2 - 1} = \frac{4n^2 - 1}{4n^2 + 8n -1}$$ $$\lim_{x\to \infty} \frac{4n^2 - 1}{4n^2 + 8n -1} = \frac{4}{4} = 1$$ Hence the test is inconclusive. How can i prove that the series is not uniformly convergent?

Answer 1

The series is uniformly convergent since for any $t$ $$\left|\frac{\cos(2nt)}{4n^2-1}\right|\leq\frac{1}{4n^2-1}$$ and $$\sum_{n=1}^{+\infty}\frac{1}{4n^2-1}=\frac{1}{2}.$$ Additionally, $$\sum_{n=1}^{+\infty}\frac{\cos(2nt)}{4n^2-1}=\frac{1}{2}-\frac{\pi}{4}\sin(t\!\!\!\!\!\pmod{\pi}).$$