Proof for Term by Term Differentiation

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.

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What do you mean a "proof using Taylor series or Laurent Series"? Have you tried differentiating it term-by-term and seeing if you get a contradiction? – user61527 Sep 15 '13 at 6:19
I will make clear what I meant to ask. We know that a Taylor series can be differentiated term by term in its disk of convergence. Therefore I was looking for a proof which could show that the given series doesn't converge. – prat Sep 15 '13 at 6:51

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.)

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