$\int_a^b\sum_{n=1}^\infty f_n(x) dx=\sum_{n=1}^\infty \int_a^b f_n(x) dx$ $\{f_n\}$ is a sequence of continuous (so also integrable) real functions on $[a,b]$ that is uniformly convergent and uniformly equicontinuous. I want to show $$\int_a^b\sum_{n=1}^\infty \frac{f_n(x)}{{\color{red}{n^2}}} dx=\sum_{n=1}^\infty \frac{1}{{\color{red}{n^2}}}\int_a^b f_n(x) dx,$$
and I cannot figure out how to do this. I think to take the limit as $n\to\infty$ and use uniform convergence to the limit $f$ which will somehow allow the sum and integrals to switch, but I cannot figure out how to do this. Help would be greatly appreciated.
Edit: In red.
 A: First note that $$\int_a^b\sum_{n=1}^{N} \frac{f_n(x)}{{{n^2}}} dx=\sum_{n=1}^{N} \frac{1}{{{n^2}}}\int_a^b f_n(x) dx$$ for each $N$ and that RHS of this equation tends to RHS of the equation we want to prove. Hence, it remains only to prove that LHS of this equation tends to LHS of the given equation. For this we use the following:
If $g_n \to g$ uniformly (with each $g_n$ continuous) then $\int_a^{b}g_n(x)dx \to \int_a^{b}g(x)dx$.
So it reamins only to prove that $\sum_{n=1}^{N} \frac{f_n(x)}{{{n^2}}}$ converges uniformly to $\sum_{n=1}^{\infty} \frac{f_n(x)}{{{n^2}}}$ as $N \to \infty$.
By M-test it is enough to show that $(f_n(x))$ is uniformly bounded. There exists $n_0$ such that $|f_n(x)-f_{n_0}(x)| <1$ for all $n \geq n_0$. Let $M$ be an upper bound for $|f_{n_0}|$ Then $|f_n(x)| <1+M$ for all $x$ for all $n \geq n_0$. Can you finish the proof now using the fact that each of the functions $f_1,f_2,...,f_{n_0-1}$ are bounded?
A: Let $f(x) = \sum\limits_{n=1}^\infty f_n(x)$. We need $$\int_a^b f(x)dx = \lim\limits_{N \to \infty} \int_a^b \sum\limits_{n=1}^N f_n(x)dx$$ Fix $\epsilon>0$, for $N$ big enough, $|f(x)-\sum\limits_{n=1}^N f_n(x)|\leq\epsilon$. Therefore $$|\int_a^b f(x)dx -\int_a^b \sum\limits_{n=1}^N f_n(x)dx| \leq \int_a^b |f(x) - \sum\limits_{n=1}^N f_n(x)|dx \leq \epsilon (b-a)$$
