Given the function $$f(x) = \sum_{n=5}^\infty \frac{1}{n^2-x^2} \text{ for } x \in (-5,5)$$ prove that it's continuous and differentiable.
So i tried doing something like $$ \sum_{n=5}^\infty \frac{1}{n^2-x^2} \le \sum_{n=6}^\infty \frac{1}{n^2-25} + \frac 1{25-x^2}. $$ The first sum is convergent, the second part is constant, so we know that our sum is uniformly convergent on the interval and that it's continuous. Similarly we can prove differentiability, is this thinking correct?
On
Are you aware of the (rather classical) result:
$$1+\sum_{n=1}^\infty \frac{2 x^2}{n^2-x^2}=\pi x \, cot(\pi x)$$
(that can be established for example using complex variable theorem of residues) or, for our purpose:
$$\sum_{n=1}^\infty \frac{1}{x^2-n^2}=\dfrac{1-\pi x \, cot(\pi x)}{2 x^2} \ \ (*)$$
This sum converges for all fixed $x \in \mathbb{R}\cap\mathbb{Z}^c$. If one removes the first 4 terms, it removes the eight poles {$\pm 1,\pm 2,\pm 3,\pm 4$} surrounding the origin, thus there is (in particular) convergence on $(-5,5)$.
Remark: in (*), the infinite values of $\cot(\pi x)$ on the RHS occur when $x$ is a positive integer, just like on the LHS.
The first term of the series is continuous and differentiable, but blows up at the ends of your interval. Pull it out.
The rest of the series will converge on your interval, and the derivatives converge uniformly. From that, the rest of the series is differentiable and its derivative is the limit of the derivatives.
Now add your troublesome first term back in. :-)