I have to find the limit:
$$\lim\limits_{n \to \infty} n \displaystyle\int_2^e (\ln x)^n dx$$
I tried using the Squeeze Theorem but it didn't work (or at least I didn't use it correctly):
$$2 \le x \le e$$
I took the log of this inequality
$$\ln(2) \le \ln(x) \le 1$$
Raised to the $n$:
$$\ln^n(2) \le \ln^n(x) \le 1$$
Took the integral from $2$ to $e$:
$$\int_2^e \ln^n(2)dx \le \int_2^e \ln^n(x) dx \le \int_2^e 1 dx$$
$$\ln^n(2) \cdot (e-2) \le \int_2^e \ln^n(x) dx \le (e-2)$$
Multiplied by n:
$$n \ln^n(2) \cdot(e – 2) \le n\int_2^e \ln^n(x) dx \le n(e-2)$$
And now, since $\ln(2) < 1$, so $\ln^n(2) = 0$ as $n \to \infty$, if I take the limit with $n \to \infty$ I get:
$$\infty \cdot 0 \le n\int_2^e \ln^n(x) dx \le \infty$$
So I didn't get anywhere trying to use the Squeeze Theorem.
Best Answer
As you said $\lim\limits_{n\to \infty}\ln^n 2=0$. Let:
$$I_n = \int_2^e\ln^n x\,dx$$
$I_n$ is decreasing because:
$$I_{n}-I_{n+1}=\int_2^e\ln^n x(1-\ln x)\,dx \geq 0$$
Integrating by parts:
$$ \begin{aligned} \displaystyle\int_2^e \ln^n x\, dx &= \displaystyle\int_2^e \ln^n x \cdot (x)'\, dx\\ &=x\ln^n x\bigg|_2^e -n\int_2^e\ln^{n-1}x\,dx\\ &= e-2\ln^n 2-nI_{n-1} \end{aligned} $$
Therefore $I_n+nI_{n-1}=e-2\ln^n2$. Now, since $I_n$ is decreasing:
$$I_n+nI_{n-1}\geq I_n+nI_n\Rightarrow I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$
and
$$I_{n+1}+(n+1)I_n\leq I_n+(n+1)I_n\Rightarrow I_n\geq \frac{1}{n+2}(e-2\ln^{n+1}2)$$
Combining the two:
$$\frac{1}{n+2}(e-2\ln^{n+1}2) \leq I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$
or
$$\frac{n}{n+2}(e-2\ln^{n+1}2) \leq nI_n\leq \frac{n}{n+1}(e-2\ln^n 2)$$
and squeezing $\lim\limits_{n\to \infty}nI_n=e$.