Showing convergence of integral sequence

Question

Suppose we have the integral $I_n = \frac{q^n}{n!}\int\limits_0^\pi x^n(\pi-x)^n \hspace{2pt} dx$ for some positive integer $q$. I'd like to prove that the sequence $\{I_n\}$ converges to zero. This is intuitively clear to me due to the factorial in the denominator, but is there a rigorous way to show this? I'd appreciate a hint.

Answer 1

Note that$$(\forall n\in\Bbb N)(\forall x\in[0,\pi]):x^n(\pi-x)^n\leqslant\left(\frac\pi2\right)^{2n}$$and that therefore$$(\forall n\in\Bbb N):q^n\int_0^\pi x^n(\pi-x)^n\,\mathrm dx\leqslant\pi\left(\frac{q\pi}2\right)^{2n}.$$So,$$(\forall n\in\Bbb N):\frac{q^n}{n!}\int_0^\pi x^n(\pi-x)^n\,\mathrm dx\leqslant\pi\frac{\left(\frac{q\pi}2\right)^{2n}}{n!}.$$

Answer 2

The antiderivative exists in terms of the gaussian hypergeometric function.

Using the given bounds $$I_n = \frac{q^n}{n!}\int\limits_0^\pi x^n(\pi-x)^n \, dx=\frac{\pi ^{2 n+1} \Gamma (n+1)}{\Gamma (2 n+2)}\,q^n$$ which converges since $$\frac{I_{n+1}}{I_n}=\frac{\pi ^2}{4 n+6}q$$

Now, using Stirling approximation and Taylor series $$I_n=\frac{\pi }{2 \sqrt{2} n} \left(\frac{e \pi ^2 }{4 n}\right)^n\exp\left(-\frac{11}{24 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)\right)\,q^n$$ which is a quite good approximation (relative error smaller than $0.1$% if $n>3$, smaller than $0.01$% if $n>7$.

Answer 3

$$I_n=\frac{q^n}{n!}\int_0^\pi x^n(\pi-x)^ndx$$ make the substitution $u=x/\pi\Rightarrow dx=\pi du$ so: $$I_n=\frac{q^n}{n!}\int_0^1(\pi u)^n(\pi-\pi u)^n\pi du=\frac{q^n\pi^{2n+1}}{n!}\int_0^1u^n(1-u)^ndu$$ $$=\frac{q^n\pi^{2n+1}}{n!}B(n+1,n+1)$$ now express the Beta function in terms of the Gamma function, and since $n$ is an integer it can then be expressed in terms of the factorial, see where this gets you :)