To what value will the series converge? I have done a Fourier series expansion and get $$\frac{12}{\pi(2n-1)}\sin((2n-1)x)$$
How to find the value it converges at $x=\frac{\pi}{2}$?
isn't it divergent?
Please show me the correct way step by step

Thanks
$$\sin((2n-1)\frac{\pi}{2})\\n=1,2,3,...\\\sin(1\frac{\pi}{2}),\sin(5\frac{\pi}{2}),\sin(3\frac{\pi}{2}),\sin(7\frac{\pi}{2})\\\frac{12}{\pi}\sum \frac{1}{(2n-1)}\sin((2n-1)\frac{\pi}{2})=\\ \frac{12}{\pi}\sum \frac{1}{(2n-1)}(-1)^n$$

and the series converge.
$$\sin((2n-1)\frac{\pi}{2})\\n=1,2,3,...\\\sin(1\frac{\pi}{2}),\sin(5\frac{\pi}{2}),\sin(3\frac{\pi}{2}),\sin(7\frac{\pi}{2})\\\frac{12}{\pi}\sum \frac{1}{(2n-1)}\sin((2n-1)\frac{\pi}{2})=\\ \frac{12}{\pi}\sum \frac{1}{(2n-1)}(-1)^n$$

and the series converge.