Showing that a series is convergent having it as an integral

Question

if $a_n = \int_0^{\pi/2} \sin^n t\,dt $

show that $a_n$ *is convergent.

Should I use the integral test to show that? if yes, please tell me how. Thank you

Answer 1

FYI,

Performing the substitutions $x=\sin t,$ then $u=x^2,$ in $\int\limits_0^{\pi/2} \sin^n t\,dt$

Firts $\frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}}$ and the limits: $0$ and $1$ we get: $\int\limits_0^{1} \frac{x^n}{\sqrt{1-x^2}} \,dx$

The second one $\frac{dx}{du}=\frac{1}{2\sqrt{u}}$ and the limits: $0$ and $1$ we get: $\frac{1}{2}\int\limits_0^1 u^{\frac{n-1}{2}}(1-u)^{-\frac{1}{2}}\,du$

Using the definition of $\beta$ function and the identity of $\Gamma$ function we have:

$a_n=\frac{1}{2} \beta(\frac{n+1}{2},\frac{1}{2})=\frac{1}{2}\frac{\Gamma(\frac{n+1}{2}) \Gamma(\frac{1}{2})}{\Gamma(\frac{n}{2}+1)}\tag1$

The simplest way to prove that $\lim\limits_{n\rightarrow\infty} a_n=0$ using Striling's approximation when ${n\rightarrow\infty}$.

$\Gamma(z+1)\approx \sqrt{2\pi z}\Big(\frac{z}{e}\Big)^z$

$ \frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2}+1)} \approx\frac{\sqrt{\pi}}{2}\frac{\sqrt{\pi (n-1)}\Big(\frac{(n-1)}{2e}\Big)^\frac{n-1}{2}}{\sqrt{\pi n}\Big(\frac{n}{2e}\Big)^{\frac{n}{2}}} =\frac{\sqrt{\pi}}{2}\sqrt{1-\frac{1}{n}}\sqrt{(1-\frac{1}{n})^n}\sqrt{\frac{2e}{n-1}}\tag2$

Finally

$\lim\limits_{n\rightarrow\infty}\frac{\sqrt{\pi}}{2}\overbrace{\sqrt{1-\frac{1}{n}}}^{\rightarrow 1}\overbrace{\sqrt{(1-\frac{1}{n})^n}}^{\rightarrow\frac{1}{\sqrt{e}}}\overbrace{\sqrt{\frac{2e}{n-1}}}^{\rightarrow 0}=0\tag3$

$\lim\limits_{n\rightarrow\infty}a_n=0\tag4$