Question about the $M$ value in Weierstrass $M$-test The Weierstrass $M$-test says that if $(f_n$) is a sequence of functions on $E$, and $\left|f_n(x)\right|\leq M_n$, then $\sum f_n$ converges uniformly on $E$ if $\sum M_n$ converges. 
My question is, can $M_n$ depend on $x$? that is, if I know $\left|f_n(x)\right|\leq \dfrac{1}{x^2n^2}$ (whose sum does converge), is this sufficient?
 A: No, $M_n$ cannot depend on $x$. That's the whole point of the M-test. In your case, you should fix a range for $x$ to obtain a bound that is independent on it. 
In other words: If you allow $x$ to be very small your bound becomes meaningless.
A: For any  $\;x\in[a,b]\;$ , we have that
$$\left|\frac x{1+n^2x^2}\right|=\frac{|x|}{1+n^2x^2}\le\frac{|b|}{1+n^2\min\{|a|,|b|\}^2}$$
and the series
$$\sum_{n=1}^\infty\frac{|b|}{1+n^2\min\{|a|,|b|\}^2}$$
converges as long as that minimum is not zero, so we have uniform convergence in such bounded intervals which don't include zero.
Thinking a second time, if $\;x=0\;$ then the function is zero so everything's fine, too.
Added: For $\;x\ge 0\;$ we have
$$f_n(x):=\frac x{1+n^2x^2}\implies f'_n(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}=\frac{(1-nx)(1+nx)}{(1+n^2x^2)^2}=0\iff x_0=\frac1n$$
and clearly we have for values of $\;x\;$ close to $\;x_0\;$
$$\begin{cases}f'_n(x)>0&,\;\;x<x_0\\{}\\f'_n(x)<0&,\;\;x>x_0\end{cases}$$
from which we get that $\;x=x_0=\frac1n\;$ is a maximal point of $\;f_n(x)\;$ on $\;[0,\infty)\;$ , and since $\;f_n(-x)=-f_n(x)\;$ the same is true in $\;(-\infty,0)\;$, and from here we get that for all $\;x\in\Bbb R\;$
$$\left|f_n(x)\right|= \left|\frac x{1+n^2x^2}\right|\le \left|f_n\left(\frac1n\right)\right|=\frac1{2n}\xrightarrow[n\to\infty]{}0\implies f_n(x)\stackrel{\text{unif. on}\;\Bbb R}{\xrightarrow[n\to\infty]{}}f(x)=0$$
