Show that the series for which the sum of first n terms
$f_n(x)=\frac{nx}{1+n^2x^2}, 0\leq x\leq 1$ can not be differentiated term by term at $x=0$. What happens at $x\neq0$.
I have found similar questions but solved using Fourier Series. I am in need of a proof using Taylor series or Laurent Series and the basics of Cauchy's complex analysis and Sequence theorems.
This seems to have very little to do with Taylor or Laurent series, but here goes:
I assume that you mean $$ \sum_{k=1}^n u_k(x) = \frac{nx}{1+n^2x^2}$$ for some implicitly defined $u_k$. (I first read the questions as if the terms of the series where given by $f_n$, but that yields a divergent series.) Then by letting $n\to \infty$, we see that $$ \sum_{k=1}^\infty u_k(x) = 0$$ for $x \in [0,1]$. If we differentiate termwise we get $$ \sum_{k=1}^n u'_k(x) = \frac{n(1-n^2x^2)}{(1+n^2x^2)^2}$$ and at $x=0$, $$ \sum_{k=1}^n u'_k(x) = n$$ which certainly doesn't tend to $0$ as $n\to\infty$. I'll leave the case $x\neq 0$ to you. (Argue using the above computations.)