I have to show that this series:
$\sum_{n=0}^n \frac{1}{n^2-x^2}$ on any closed interval that dosnt contain an integer.
converges uniformly and absolutely.
I know that $\frac{1}{n^2-x^2}$ $\le$ $\frac{1}{n^2C}$ and how to show that it converges uniformly/absolutely from the M-test and the p-test.
I am having trouble though with how it was found that $\frac{1}{n^2-x^2}$ $\le$ $\frac{1}{n^2C}$
Could someone people help me understand how this was found to be the case?
For large enough $n$, we have
$$0\leq x\leq C \implies$$ $$ n^2-C^2\leq n^2-x^2 \leq n^2$$
$$\implies 0<\frac{1}{n^2-x^2}\leq \frac{1}{n^2-C^2}$$
but $$\frac{1}{n^2-C^2}\sim \frac{1}{n^2}\;(n\to+\infty)$$
thus
$\sum\frac{1}{n^2-x^2}$ is normally and uniformly convergent at $[0,C]$.