What is uniform convergence for one function and how is that equivalent to continuity? My teacher wanted to prove that function
$$\frac{1}{x} - \sum_{n=1}^\infty \frac{2x}{n^2-x^2}$$
is continuous. He said that it is equivalent to proving that
$$\sum_{n=1}^\infty \frac{1}{n^2-x^2}$$
uniformly converges. However uniform convergence is defined for a sequence of function as I understand it. Here I see only one function: the equation depends purely on x and nothing else than x - that is just one function in terms of x.
But even if it made sense, how is uniform convergence equivalent to continuity?
According to my knowledge he didn't even prove uniform convergence (despite saying that). He only proved that the equation above:
$$\sum_{n=1}^\infty \frac{1}{n^2-x^2}$$
converges (to a number smaller than something). And convergence (as I understand it) is different from uniform convergence. 
How is that correct?

A little addendum: not only such function is continuous on its maximal domain $\mathbb{R}\setminus\mathbb{Z}$, but it is a well-known function. If we start with the Weierstrass product for the sine function
$$ \frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)\tag{1} $$
and apply $\frac{d}{dx}\log(\cdot)$ to both sides we get:
$$ \sum_{n\geq 1}\frac{2x}{x^2-n^2} = \pi\cot(\pi x)-\frac{1}{x}\tag{2} $$
hence:
$$ \frac{1}{x}-\sum_{n\geq 1}\frac{2x}{n^2-x^2} = \color{red}{\pi\cot(\pi x)}.\tag{3} $$
