Compute $\lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx$ 
Compute $$\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx.$$

I tried writing $\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx = \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{1-(\sin{x})^{n+1}}{1-\sin{x}}dx$, but I don't know how to continue form here.
I know that the limit diverges to $\infty$, but I don't know how to prove it.
 A: The wanted limit is $+\infty$. Indeed, by letting $I_k=\int_{0}^{\pi/2}\left(\sin x\right)^k\,dx$ we have that $\{I_k\}_{k\geq 1}$ is a decreasing sequence, but it is also log-convex by the Cauchy-Schwarz inequality, and by integration by parts
$$ I_{k}^2\geq I_k I_{k+1} = \frac{\pi}{2(k+1)}\tag{1}$$
such that
$$ \sum_{k=1}^{n}I_k \geq \sqrt{\frac{\pi}{2}}\sum_{k=1}^{n}\frac{1}{\sqrt{k+1}}\geq \sqrt{\frac{\pi}{2}}\sum_{k=1}^{n}2\left(\sqrt{k+2}-\sqrt{k+1}\right)=\sqrt{2\pi}\left(\sqrt{n+2}-\sqrt{2}\right).\tag{2}$$
A: From the inequality $\cos x\geq 1-\frac{x^2}{2}$, we have that $$\sin x=\cos\left(x-\frac{\pi}{2}\right)\geq 1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2.$$ So, from Bernoulli's inequality, for $x$ close to $\pi/2$, $$(\sin x)^k\geq\left(1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2\right)^k\geq 1-\frac{k}{2}\left(x-\frac{\pi}{2}\right)^2.$$ Therefore, for $\delta>0$ small, $$\int_{\pi/2-\delta}^{\pi/2}(\sin x)^k\,dx\geq\int_{-\delta}^0\left(1-\frac{k}{2}y^2\right)\,dy=\delta-\frac{k\delta^3}{6},$$ and choosing $\delta=1/\sqrt{k}$ we obtain that $$\int_{\pi/2-1/\sqrt{k}}^{\pi/2}(\sin x)^k\,dx\geq\frac{1}{\sqrt{k}}-\frac{1}{6\sqrt{k}}=\frac{5}{6\sqrt{k}},$$ which shows that the limit is $\infty$.
