# Show that for all $n \in \Bbb{N}^*$, $\frac{1}{n+1} \le u_n$ [closed]

$$u_n:= \int_{1}^{e}x^{1/3}(1-\ln(x))^n \, \mathrm{d}x$$

1. I showed that $$u_n$$ is decreasing.
2. I showed that for all $$n \in \Bbb{N}^*$$, $$u_{n+1} = -\dfrac{3}{4} + \dfrac{3}{4}(n+1)u_n$$.
3. I have to show that for all $$n \in \Bbb{N}^*$$, $$\dfrac{1}{n+1} \le u_n.$$

Do you have any idea? I tried to use the fact that the sequence is decreasing but it doesn't help.

Note that $$0\le\ln x\le 1$$ for $$1\le x\le e$$, hence $$x^{1/3}(1-\ln x)^n\ge0$$ on the interval of integration, and thus $$u_n\ge0$$ for all $$n$$. If you've shown that $$u_{n+1}=-{3\over4}+{3\over4}(n+1)u_n$$, then $$u_n\ge{1\over n+1}$$ follows from the observation that $$u_{n+1}\ge0$$.

Taking the $$Z-$$transform for the equation that you proved, i.e. $$$$u_{n+1}=-\frac{3}{4}+\frac{3}{4}(n+1)u_n$$$$ $$$$zU(z)-zu(0)=-\frac{3}{4}\frac{z}{z-1}-\frac{3z}{4}\frac{\text{dU(z)}}{\text{dt}}+\frac{3}{4}U(z)$$$$ Assuming $$u(0)=1$$, and after some manipulation of the above equation and taking inverse $$Z-$$transform, we get, $$$$u_n=\frac{1}{4}\Bigg((4c_1+3)\Big(\frac{3}{4}\Big)^n\Gamma(n+1)-4e^\frac{4}{3}E_{-n}\Big(\frac{4}{3}\Big)\Bigg)$$$$ To find $$c_1$$, we substitute $$u_0=1$$ and use the asymptotic expansion of $$E_n(x)$$ viz., $$$$E_n(x)= \frac{e^{-x}}{x}\Bigg[1-\frac{n}{x}+\frac{n(n+1)}{x^2}-...\Bigg]$$$$ $$$$E_0\Big(\frac{4}{3}\Big)=\frac{3}{4}e^{-\frac{4}{3}}$$$$ Thus, $$c_1=1$$ Thus, $$$$u_n=\frac{1}{4}\Bigg(7\cdot\Big(\frac{3}{4}\Big)^n\Gamma(n+1)-4e^{\frac{4}{3}}E_{-n}\Big(\frac{4}{3}\Big)\Bigg).$$$$ Now, replacing $$n$$ with $$-n$$ and $$x$$ by $$\frac{4}{3}$$ in the Exponential Integral function $$E_n(x)$$, we get, $$$$u_n=\frac{1}{4}\Bigg(7\cdot\Big(\frac{3}{4}\Big)^nn!-3\displaystyle \sum_{k=0}^{n} \Bigg(\frac{3}{4}\Bigg)^k{}^nP_k\Bigg)$$$$ This can be represented in other form as: $$$$y_n=\frac{4^{-n}\Big(7\cdot 3^{n+1}\cdot (n+1)!-4^{n+1}e^{4/3}E_{-n}\Big(\frac{4}{3}\Big)(n+3)\Big)}{3(n+1)}$$$$ where, $$y_n = 4u_n-\frac{4}{n+1}$$. Since this is a strictly increasing function, it is proved that $$u_n\ge\frac{1}{n+1}$$

Wolframalpha plot for $$y_n$$

The map

$$f(x) = 1-\ln x$$ is convex, $$f(1)= 1$$ and $$f(e)=0$$

Hence $$f(x)$$ is above its tangent at $$1$$ which is the line $$y=2-x$$.

From this, we get

$$u_n= \int_{1}^{e}x^{1/3}(1-ln(x))^n dx\ge \int_1^e (2-x)^n \ dx \ge \int_1^2 (2-x)^n \ dx =\frac{1}{n+1}$$