Analysis: Show that $f(x)$ converges, pointwise and uniformly on $\mathbb{R}$ to a differentiable function $f$ that satisfies: Show that
$$f(x) = \sum_{k=1}^\infty \frac{\sin kx}{k^3} \tag1$$
converges, pointwise and uniformly on $\mathbb R$, to a differentiable function that satisfies
$$\int_0^{\pi/2} f(x)\,dx = \sum_{k=1}^\infty \left(\frac{(-1)^{k-1}}{(2k)^4}+\frac{1}{k^4}\right)\tag{a}$$
and 
$$
|f'(x)|\le \pi/8 \tag{b}
$$
I would be able to do it if it wasn't for the part "$f$ that satisfies" (a) and (b).  If someone could explain it to me, that would be helpful.
 A: The question is to prove that the function 
$$f(x) = \sum_{k=1}^\infty \frac{\sin kx}{k^3} \tag1$$
satisfies 
$$\int_0^{\pi/2} f(x)\,dx = \sum_{k=1}^\infty \left(\frac{(-1)^{k-1}}{(2k)^4}+\frac{1}{k^4}\right)\tag{a}$$
and 
$$
|f'(x)|\le \pi/8 \tag{b}
$$

Part (a) follows from uniform convergence of the series (1), which allows us to exchange integration and summation: 
$$\int_0^{\pi/2} f(x)\,dx = \sum_{k=1}^\infty \int_0^{\pi/2} \frac{\sin kx}{k^3} \,dx
=\sum_{k=1}^\infty \left(\frac{\cos 0}{k^4}-\frac{\cos k\pi/2}{k^4}\right)
\tag2$$
which is (a).
Part (b) requires the consideration of the series of derivatives:
$$g(x) = \sum_{k=1}^\infty \frac{\cos kx}{k^2} \tag3$$
Since (3) converges uniformly, it follows that $f$ is an antiderivative of $g$. Put another way, $f$ is differentiable and $f'=g$. Since $g(0)=\sum_{k=1}^\infty 1/{k^2}=\pi^2/6 > \pi/8$, inequality (b) is false. 
The sharp upper estimate for $|f'|$   is $|f'(x)|\le \pi^2/6$, obtained by using $|\cos kx|\le 1$  in (3). 

By the way, 
$$f(x)=\frac{1}{12}(x^3-3\pi x^2+2\pi^2 x), \quad 0<x<\pi$$
