I tried to create a infinite series which can not be differentiated by parts. I took $f_N(x)=\sum_{n=1}^N (\frac{1}{2})^n\frac{\sin(nx)}{n}$ with $\sum_{n=1}^\infty (\frac{1}{2})^n\frac{\sin(nx)}{n}=e^{\frac{1}{2}\cos(x)}\sin(\frac{1}{2}\sin(x))$. Derivating the series by parts we get $f_N'(x)=\sum_{n=1}^N(\frac{1}{2})^n\cos(nx)$ with $\sum_{n=1}^\infty(\frac{1}{2})^n\cos(nx)=\frac{1-\frac{1}{2}\cos(x)}{1-\cos(x)+\frac{1}{4}}-1$. Now taking the derivative of $e^{\frac{1}{2}\cos(x)}\sin(\frac{1}{2}\sin(x))$ we get $e^{\frac{1}{2}\cos(x)}(-\frac{1}{2}\sin(x))\sin(\frac{1}{2}\sin(x))+e^{\frac{1}{2}\cos(x)}\cos(\frac{1}{2}\sin(x))\frac{1}{2}\cos(x)$ what is not the same as $\frac{1-\frac{1}{2}\cos(x)}{1-\cos(x)+\frac{1}{4}}-1$. Just what i was looking for. A picture to proof this: enter image description here
But now my problem: I can not argue why the derivative of $f_N(x)$ differs the derivative of the limit of $f_N(x)$ since both of the series $f_N(x)$ and $f'_N(x)$ converege uniformly due to the M-test ($|\sum_{n=1}^\infty (\frac{1}{2})^n\frac{\sin(nx)}{n}|\leq \sum_{n=1}^\infty\frac{1}{2^n}$ and $|\sum_{n=1}^\infty(\frac{1}{2})^n\cos(nx)|\leq\sum_{n=1}^\infty\frac{1}{2^n}$). I thought the uniformly convergence of both series allows us to differentiate by parts. Does anybody know where I am wrong?