# Why do this series converge to 2 different values?

Firstly, I consider the Fourier series of the $$2\pi$$-periodic function $$f=\cos{\frac{3}{2}\pi}$$ if $$-\pi\leq x\leq \pi.$$ And I get $$\cos\frac{3}{2}t \sim -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nt.$$ Since I find that $$f\in C^2(T),$$ i.e. $$f,f’$$ and $$f’'$$ are continuous on $$T,$$ then $$\cos\frac{3}{2}t = -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nt \text{ (converge uniformly)}.$$ Let $$x=\frac{\pi}{2},$$ then \begin{align} \cos\frac{3}{2}(\frac{\pi}{2})&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos n\frac{\pi}{2}\notag\\ &=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag \end{align} Let $$x=\frac{3}{2}\pi,$$ then \begin{align} \cos\frac{3}{2}(\frac{3}{2}\pi)&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos n\frac{3}{2}\pi\notag\\ &=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag \end{align} I find that these two same series $$-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}$$ converge to 2 different values, which are $$\cos\frac{3}{2}(\frac{\pi}{2}) \text{ and } \cos\frac{3}{2}(\frac{3}{2}\pi).$$ Did I make any mistake? Thanks~

No, but your function is not $$cos(\frac{3x}{2})$$ as such, it is a periodic extension of that defined in $$-\pi$$ to $$\pi$$. If we call this function $$f(x)$$ then you can see in the graph below that $$f(\frac{\pi}{2})$$ and $$f(\frac{3 \pi}{2})$$ are the same.