Show that for all $n \in \Bbb{N}^*$, $\frac{1}{n+1} \le u_n$ $$u_n:= \int_{1}^{e}x^{1/3}(1-\ln(x))^n \, \mathrm{d}x$$


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*I showed that $u_n$ is decreasing.

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

*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.
 A: Taking the $Z-$transform for the equation that you proved, i.e.
\begin{equation}
u_{n+1}=-\frac{3}{4}+\frac{3}{4}(n+1)u_n
\end{equation}
\begin{equation}
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)
\end{equation}
Assuming $u(0)=1$, and after some manipulation of the above equation and taking inverse $Z-$transform, we get,
\begin{equation}
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)
\end{equation}
To find $c_1$, we substitute $u_0=1$ and use the asymptotic expansion of $E_n(x)$ viz.,
\begin{equation}
E_n(x)= \frac{e^{-x}}{x}\Bigg[1-\frac{n}{x}+\frac{n(n+1)}{x^2}-...\Bigg]
\end{equation}
\begin{equation}
E_0\Big(\frac{4}{3}\Big)=\frac{3}{4}e^{-\frac{4}{3}}
\end{equation}
Thus, $c_1=1$
Thus,
\begin{equation}
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).
\end{equation}
Now, replacing $n$ with $-n$ and $x$ by $\frac{4}{3}$ in the Exponential Integral function $E_n(x)$, we get,
\begin{equation}
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)
\end{equation}
This can be represented in other form as:
\begin{equation}
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)}
\end{equation}
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 $y_n$
A: 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 $u_{n+1}\ge0$.
A: 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 
$$u_n= \int_{1}^{e}x^{1/3}(1-ln(x))^n dx\ge \int_1^e (2-x)^n \ dx \ge \int_1^2 (2-x)^n \ dx =\frac{1}{n+1}$$
