How does $\sum _n (-1)^n \int_0^{\frac{\pi }{2}} \cos ^n(x) \, dx$ converge??

Question

In my last question: Does $\sum _n \int_0^{\frac{\pi }{2}} \cos ^n(x) \, dx$ converge? I've managed to transform the expression using: https://en.wikipedia.org/wiki/Monotone_convergence_theorem

Could sb help me just with this transformation in a bit more complex example? How may I treat $(-1)^n$ inside the sum?

Answer 1

For $\frac{\pi}{2}\ge x\gt 0$, $\sum_{n=0}^\infty (-cos(x))^n=\frac {1}{1+cos(x)}$. For $x=0$, the sum does not converge but the partial sums are bounded , either $=0$ or $=1$, so the integral is not a problem. I'll leave to you to get$\int_0^{\frac{\pi}{2}} \frac{1}{1+cos(x)}dx$.

Answer 2

You do not need any closed formula here if you are just interested in convergence/divergence.

$0< \cos x< 1$ on $(0,\pi/2)$. Therefore, $\cos^{n+1}x<\cos^nx$ on that same interval. By monotonicity of the integral, you also have $$ \int\limits_0^{\pi/2}\cos^{n+1}x\,dx<\int\limits_0^{\pi/2}\cos^{n}x\,dx. $$ Moreover, $\cos^nx\downarrow 0$ as $n\to\infty$ monotonically on $(0,1)$ and, consequently (monotonic convergence) $$ \int\limits_0^{\pi/2}\cos^{n}x\, dx\to 0. $$ Then it just remains to apply the Leibnitz test (Alternating series test) to conclude that your series is convergent.