<s>[INST] 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.
 [/INST] 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 $I_N$. 
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