Given $g_n(x)= \frac{nx}{1+n^2 x}$, $g_n : [0,1] \to \mathbb R, n \in \mathbb N, n\geq 1$, I am trying to show that $g_n$ converges uniformly. I have shown that it converges pointwise to $g:[0,1]\to\mathbb R$, with $g(x)=0$. Now I wish to show that it converges uniformly.
I know how to do this using the epsilon definition, however I tried applying the M test, which states that if $$\sum_{n=1}^{\infty} ||g_n||_\infty < \infty$$ then the sequence converges uniformly. However, for this sequence, this yields me with $$\sum_{n=1}^\infty ||\frac{nx}{1+n^2x}||_\infty =\sum_{n=1}^\infty \sup_{x\in [0,1]} \left|\frac{nx}{1+n^2x}\right|=\sum_{n=1}^\infty \left|\frac n {1+n^2} \right|.$$ To my knowledge, this is divergent.
On the contrary, we were told that $$\lim_{n\to \infty} ||f_n - f||_\infty = 0 \iff f\text{ is uniformly convergent.}$$ Using this definition here, however, yields $$\lim_{n\to\infty}||\frac{nx}{1+n^2x}-0||_\infty = \lim_{n\to\infty}\frac{n}{1+n^2}=0.$$
I have tried searching yet am unable to clarify where my mistake here lies. Is the M test not an equivalence? Or am I merely mistaken?
If I understand correctly you are asking if the following implication holds: $$(g_n) \ \text{converges uniformly} \implies\Sigma||g_n||_\infty<\infty $$
To see that this is false take $\forall x\ \ :g_n(x)=\frac{1}{n}$.