Squeeze an integral Would you have any idea about this problem ?

Prove that for all nonnegative integers $n$, the following inequalities hold:
  $$\frac{e^2}{n+3}\leq \int_{1}^{e} x (\ln x)^n \,dx \leq \frac{e^2}{n+2} ,$$ 
  where $\ln$ is the natural logarithm.

Thank you.
 A: The rightmost inequality can be proved using substitution and integration by parts.
In fact, set $I(n):=\int_1^e x\ (\ln x)^n\ \text{d} x$; then:
$$I(n) \stackrel{t=\ln x}{=} \int_0^1 e^{2t}\ t^n\ \text{d} t$$
and integrating by parts three times the RH side yields:
$$\tag{1} I(n)= \frac{e^2 (n^2+3n+4)}{(n+1)(n+2)(n+3)} - \frac{8}{(n+1)(n+2)(n+3)}\ I(n+3)\; ;$$
since $I(\cdot) \geq 0$ and $n^2+3n+4\leq (n+1)(n+3)$, you get:
$$I(n)\leq \frac{e^2 (n^2+3n+4)}{(n+1)(n+2)(n+3)} \leq \frac{e^2}{n+2}\; .$$
I didn't make the computations, but I bet your leftmost inequality can be proved in analogous manner... It seems all you have to do is another integration by parts in (1).

Following Didier advice, I complete the answer.
Once you got the the upper bound for $I(n)$, you have two ways to obtain the desidered lower bound for $I(n)$.


*

*Didier's way: use the upper bound for $I(n+3)$ to lower bound $I(n)$. 
Since $I(n+3)\leq \frac{e^2}{n+5}$, from (1) and $8/(n+5)\leq 2$ you gain:
$$\begin{split}I(n) &\geq \frac{e^2 (n^2+3n+4)}{(n+1)(n+2)(n+3)} - \frac{8}{(n+1)(n+2)(n+3)}\ \frac{e^2}{n+5}\\ &\geq \frac{e^2(n^2+3n+4)}{(n+1)(n+2)(n+3)} -\frac{2e^2}{(n+1)(n+2)(n+3)} \\ &= \frac{e^2 (n^2+3n+2)}{(n+1)(n+2)(n+3)}\\ &\geq \frac{e^2}{n+3}\; .\end{split}$$

*Pacciu's way: use again integration by parts. 
In fact, another integration by parts in (1) yields:
$$I(n)=\frac{e^2(n^3+7n^2+16n+8)}{(n+1)(n+2)(n+3)(n+4)}+\frac{16}{(n+1)\cdots (n+4)}\ I(n+4)$$
therefore you can use $I(\cdot )\geq 0$ and $n^3+7n^2+16n+8 \geq (n+1)(n+2)(n+4)$ to get your lower bound.
A: If $I(n)=\int _1^ex (\ln (x))^ndx$, then by integrating by parts we have $I
(1)=\frac{1}{4} \left(1+e^2\right)$ and $I(n)=\frac{e^2}{2}-\frac{n}{2}\times I(n-1)$ for $n>1$.Now your inequality can be verified by induction on $n$
A: As bleh notes already, if $I_n = \int_1^e x \cdot (\ln x)^n \, dx$, then integration by parts gives
$$
I_n = \frac{e^2}{2} - \frac{n}{2} I_{n-1}.
$$
Now, multiplying the above equation by $\frac{(-2)^n}{n!}$, we get
$$
\frac{(-2)^n  I_n}{n!} = \frac{e^2}{2} \cdot \frac{(-2)^n}{n!} + \frac{(-2)^{n-1}}{(n-1)!} I_{n-1}.
$$
Plugging in $n = 1, 2, \ldots, N$, and adding the $N$ equations, we get 
$$
\frac{(-2)^N  I_N}{N!} = \frac{e^2}{2} \cdot \sum_{n=1}^N \frac{(-2)^n}{n!} + \frac{(-2)^{0}}{0!} I_{0}, \tag{$\ast$}
$$
after a telescopic cancellation. Now $I_0 = \int_1^e x \, dx = \frac{e^2}{2} - \frac{1}{2} = \frac{e^2}{2} (1 - e^{-2})$. Plugging this in $(\ast)$, we get
$$
\frac{(-2)^N  I_N}{N!} = \frac{e^2}{2} \cdot \left( \sum_{n=0}^N \frac{(-2)^n}{n!} - e^{-2} \right). \tag{$\ast\ast$}
$$
Using the series expansion $e^{-2} = \sum \limits_{n=0}^{\infty} \frac{(-2)^n}{n!}$, we can rewrite $(\ast\ast)$ as
$$
\frac{(-2)^N  I_N}{N!} = - \frac{e^2}{2} \cdot \sum_{n=N+1}^{\infty} \frac{(-2)^n}{n!} ,
$$
or
$$
I_N = e^2 \left( \frac{1}{N+1} - \frac{2}{(N+1)(N+2)} + \frac{4}{(N+1)(N+2)(N+3)} - \cdots \right),
$$
which is an alternating series whose terms are monotone decreasing (in magnitude). Therefore, the odd and even partial sums provide upper and lower bounds on $I_N$. 
