I'm not sure how differentiability works with uniform convergence. My book says that we can show this (calculation wise) $$\varepsilon (x,a) = \sum_{k=1}^{\infty} E_{k}(x,a)$$
for some $x$ and $a$.
In this context, I'm not sure how to go about doing this:
Prove that $$f(x)=\sum_{n=1}^{\infty}\frac{1}{n^{2}+x^{2}}$$ is differentiable for all values of $x$.
If someone could give me a hand, that would be great. My book has no examples in this topic, so I just chose one of the first questions from it.
(If more context is needed, please let me know. This is my first stackexchange question.)
You cannot always differentiate term-by-term but in this case we can. Let $f_n(x)=1/(n^2+x^2)$, let $F_k(x)=\sum_{n=1}^kf_n(x)$ , let $$f(x)=\sum_{n=1}^{\infty}f_n(x)=\lim_{k\to \infty}F_k(x).$$ Let $g_k(x)=\sum_{n=1}^kf'_n(x)$ and let $$g(x)=\sum_{n=1}^{\infty}f'_n(x)=\lim_{k\to \infty}g_k(x).$$ Now each $g_k$ is continuous. If you can show that $g_k$ converges uniformly to $g$ on any bounded real set as $k\to \infty$, then you know $g$ is continuous and that $$f(x)-f(a)=\lim_{k\to \infty}(F_k(x)-F_k(a))=\lim_{k\to \infty}\int_a^xg_k(y) dy=\int_a^xg(y)dy$$ for any $x,a.$ Now since $g$ is continuous we have (with any fixed $a\ne x$),$$f'(x)=g(x).$$