I am trying to make sure I am using the Weierstrass test correctly. So, on the following expression: $$f_n(x) = \frac{x}{1+nx^2}, -1 \le x \le 1$$
I looked at it this way. First I checked to see what happens at -1 and 1. The "plug in" test gets me a limit of 1/2 at x=1 and n=1, and a limit of 1/2 at x=-1 and n=1. All-righty then, we do a little L'Hôpital's rule: $$\lim_{n \rightarrow \infty} f_n(x) = \frac{\frac{d}{dn}(x)}{\frac{d}{dn}(1+ nx^2)}= \frac {0}{x^2}=0 $$
So I know this converges to zero. But does it do so uniformly? Well, I can see that this function has a bound at x=1/2. So I will use $\frac{1}{1+n}$ as my M-function. Giving us: $M_k = \frac{1}{1+n},$ and $\sum_{n=1}^{\infty}\frac{1}{1+n}$.
$\frac{1}{1+n}$ is always going to be smaller than $\frac{x}{1+nx^2}$ for any n with a particular x. That's becaue x will be between -1 and 1 and $x^2$ is therefore going to be less than 1, so (as someone pointed out below, thanks!) $1+nx^2$ is always going to be less than $1+n$. So we can say $$ \left| \frac{x}{1+nx^2} \right| \ge \frac{1}{1+n}$$ for all n. In addition to that, $\frac{1}{1+n} \lt \infty$ for all n. So we DO NOT have uniform convergence.
Did I get it right?
Thanks!