prove a sequence of functions that are partial sums of a series converges let 
$f_n:[-\frac{\pi}{2},\frac{\pi}{2}]\to \mathbb{R}$
be defined by 
$\displaystyle f_n(x):=\sum_{k=1}^n \left(\frac{x}{2}\right)^k\!\sin(kx)$. 
Prove that the sequence $\{f_n\}_{n=1}^\infty$ is uniformly convergent.
My attempt:
$\sum_{k=1}^n|(\frac{x}{2})^k\sin(kx)| \leq \sum_{k=1}^n|(\frac{x}{2})^k| \;$  since $ |\sin(x)| \leq 1. \\ $ 
 This is geometric series and $|\frac{x}{2}| \leq 1$ on this interval, so  $\sum_{k=1}^n|(\frac{x}{2})^k|$ is convergent for all $x \in [-\frac{\pi}{2},\frac{\pi}{2}]$, and it follows that 
$\sum_{k=1}^n|(\frac{x}{2})^k\sin(kx)|$  is convergent and therefore $\sum_{k=1}^n(\frac{x}{2})^k\sin(kx)$ is convergent.
writing out some of the functions of the sequence:
$f_1: (\frac{x}{2})\sin(x) \\f_2: (\frac{x}{2})\sin(x) + (\frac{x^2}{2^2})\sin(2x)  \\f_3: (\frac{x}{2})\sin(x) + (\frac{x^2}{2^2})\sin(2x) + (\frac{x^3}{2^3})\sin(3x)\\f_n: (\frac{x}{2})\sin(x) + (\frac{x^2}{2^2})\sin(2x) + (\frac{x^3}{2^3})\sin(3x) + \cdots +  (\frac{x^n}{2^n})\sin(nx) \\$
as $n$ appraoches infinity,  the functions are converging to the infinite sum of $\sum_{k=1}^\infty \left(\frac{x}{2}\right)^k\!\sin(kx)$, and since $\sum_{k=1}^\infty \left(\frac{x}{2}\right)^k\!\sin(kx)$ is convergent, the sequence $\{f_n\}_{n=1}^\infty$ is a Cauchy sequence.
Let $S(x) = \lim_{n \to \infty} f_n(x)$
$|S(x) - f_n(x)| =  \lim_{k \to \infty}|\frac{x^{n+1}}{2^{n+1}}\sin((n+1)x) + \frac{x^{n+2}}{2^{n+2}}\sin((n+2)x) + \cdots + \frac{x^{n+k}}{2^{n+k}}\sin((n+k)x)|< \epsilon \text{ because it is a convergent series}$ 
I was also trying to apply Weierstrass M-test for this, but the m-test seems to apply for an infinite series of functions, but for this question it is a sequence of functions that are all partial sums of a series, so I was not sure how to apply it. 
 A: $|(\frac x  2)^{k} \sin (kx)| \leq (\frac {\pi} 4)^{k}$ and $\frac {\pi} 4 <1$ so M-test applies. 
A: You wrote:

since $\sum_{k=1}^\infty \left(\frac{x}{2}\right)^k\!\sin(kx)$ is convergent, the sequence $\{f_n\}_{n=1}^\infty$ is a Cauchy sequence.

From the (pointwise rather than uniform) convergence of that series, you can conclude that for every value of $x\in[-\pi/2,\pi/2]$ separately the sequence $\{f_n(x)\}_{n=1}^\infty$ is a Cauchy sequence. But what does it mean to say the sequence $\{f_n\}_{n=1}^\infty,$ without the $x,$ is a Cauchy sequence? That can make sense only if you have some notion of distance between two functions.
Let $d(f,g) = \sup\{ |f(x)-g(x)| : -\pi/2\le x\le \pi/2 \}.$


*

*Show that $d$ satisfies what is required by the definition of a metric.

*Show that $d(f_n,f)\to0$ as $n\to\infty$ if and only if $f_n\to f$ uniformly (not only pointwise) as $n\to\infty.$
Now show that if the sequence of functions is a Cauchy sequence with respect to this metric, then for every $x$ separately the sequence of values of the function at $x$ is a Cauchy sequence. That proves pointwise convergence of the sequence of functions. Next, remember that a sequence of continuous functions can converge uniformly only to a continuous function. Thus the space of continuous functions with this metric $d$ is a complete metric space.
To show the sequence of present interest is a Cauchy sequence with this metric, think about the distance between $x^m\sin(mx)$ and $x^n\sin(mx)$ and how to bound this by using $|\sin(\text{anything})|\le 1.$
