real analysis - 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.

Answer

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
$$\sum_{k=1}^{n}I_k \geq \sqrt{\frac{\pi}{2}}\sum_{k=1}^{n}\frac{1}{\sqrt{k+1}}\geq \sqrt{\frac{\pi}{2}}\sum_{k=1}^{n}2\left(\sqrt{k+2}-\sqrt{k+1}\right)=\sqrt{2\pi}\left(\sqrt{n+2}-\sqrt{2}\right).\tag{2}$$