The problem that's needed to be solved is:
Calculate $\lim_{n\rightarrow\infty} \int_{0}^{\pi/2} \frac{dx}{\sum_{k=0}^{n}(\frac{\sin x}{2})^k}$
My attempt:
Calculating this limit will probably be easier if we somehow could move our limit inside our integral. But in order to do this, one has to show that $u_n(x) := \sum_{k=0}^{n}(\frac{\sin x}{2})^k$ converges uniformly on $[0,\pi/2]$.
We know that since $|\sin(x)/2| \leq 1 \forall x$, the summation can be rewritten to:
$$u_n(x) = \frac{1-(\frac{\sin x}{2})^n}{1-\frac{\sin x}{2}}$$
since we have a geometric summation for a fixed $x$. Now, let $n \rightarrow \infty$, we get that $\lim_{n\rightarrow \infty} u_n(x) := u(x) = \frac{1}{1-\sin(x)/2}$.
Our hypothesis is that $u(x)$ is our limiting function. Using the definition of uniform convergence, we have that:
$$\sup_{x\in[0,\pi/2]} |\frac{1-(\frac{\sin x}{2})^n}{1-\frac{\sin x}{2}} - \frac{1}{1-\sin(x)/2}|= \frac{1}{1-\frac{1}{2}} - \frac{1}{1-\frac{1}{2}} \rightarrow 0, n\rightarrow \infty$$
Hence, we have shown that $u(x) $ is our limiting function, and that $u_n(x)$ converges uniformly to this function $\forall x \in [0,\pi/2]$ for large $n's$.
Thus, we can move our limit inside the integrand, and we get the following integral:
$$\int_{0}^{\pi/2} (1-\frac{\sin(x)}{2}) dx = x + \frac{\cos(x)}{2} |_{0}^{\pi/2} = \pi/2 - 1/2$$
Which is the sought limit.
I'd be glad if some of you could return any feedback on whether my solution is correct and what things that can be improved. Thanks!
I think your answer is mostly good, two suggestions
It is important that $\vert \sin(x)/2\vert < 1$ for each $x\in[0,\pi/2]$ in order to be able to sum the series.
In your argument for uniform convergence you have basically taken $n\to\infty$ and then say the result goes to $0$ as $n\to \infty$. Instead bound it for fixed $n$ independent of $x$ such as
$$\begin{align}\left\vert\frac{1-\left(\frac{\sin(x)}{2}\right)^n}{1-\frac{\sin(x)}{2}} -\frac{1}{1-\frac{\sin(x)}{2}}\right\vert&=\left\vert\frac{\left(\frac{\sin(x)}{2}\right)^n}{1-\frac{\sin(x)}{2}}\right\vert\\&\leq2\left\vert\left(\frac{\sin(x)}{2}\right)^n\right\vert\\ &\leq2\left(\frac{1}{2}\right)^n\to 0\end{align}$$