Verify the proof that $x_n = \ln^2(n+1) - \ln^2n$ is a bounded sequence. 
Let $n\ \in \mathbb N$ and:
  $$
x_n = \ln^2(n+1) - \ln^2n
$$
  Prove that $x_n$ is a bounded sequence.

I've taken the following steps. Consider $x_n$
$$
\begin{align}
x_n &= \ln^2(n+1) - \ln^2n = \\ 
&= (\ln(n+1) + \ln n)(\ln (n+1) - \ln n) = \\
&= \ln \frac{n + 1}{n}\cdot \ln (n(n+1)) = \\
&= \ln\left({1 + {1\over n}}\right)\cdot \ln(n(n+1))
\end{align}
$$
Now multiply and divide by $n$:
$$
\begin{align}
x_n &= {n \over n} \ln\left({1 + {1\over n}}\right)\cdot \ln(n(n+1)) = \\
&= \ln\left({1 + {1\over n}}\right)^n \cdot\ln \sqrt[^n]{(n(n+1))}
\end{align}
$$
Now consider $\left({1 + {1\over n}}\right)^n$. There are plenty of proofs that it is bounded. In my case I've used expansion with binomial coefficients to prove that :
$$
2< \left({1 + {1\over n}}\right)^n < 3 \implies \\
\ln2 < \ln \left({1 + {1\over n}}\right)^n < \ln3
$$
So now we want to prove that $\ln \sqrt[^n]{(n(n+1))}$ is bounded. Start with the following:
$$
\ln \sqrt[^n]{n(n+1)} < \ln \sqrt[^n]{(n+1)^2}
$$
Consider the following equation:
$$
\begin{align}
\sqrt[^n]{(n+1)^2} &= 1+a_n \iff \\
\iff (n+1)^2 &= (1+a_n)^n = \sum_{k=0}^{n}\binom{n}{k}a_n^k
\end{align}
$$
Now:
$$
\sum_{k=0}^{n}\binom{n}{k}a_n^k \ge \frac{n(n+1)}{2}a_n^2 \implies \\
\implies (n+1)^2 \ge \frac{n(n+1)}{2}a_n^2 \implies \\
\implies a_n \le \sqrt{2 + {2\over n}}
$$
So $a_k$ is clearly bounded. Which means:
$$
\sqrt[^n]{(n+1)^2} < 1 + \sup\{a_n\} = 3
$$
Also $\sqrt[^n]{(n+1)^2} > 1$. So:
$$
\ln1 < \ln \sqrt[^n]{(n(n+1))} < \ln3
$$
Now going back to initial expression:
$$
\ln1 \cdot \ln2 < \ln\left({1 + {1\over n}}\right)^n \cdot\ln \sqrt[^n]{(n(n+1))} < \ln3 \cdot \ln3
$$
Meaning $x_n$ is bounded. Have I missed something?
 A: It's well done. I just want to observe that it is rather easy to prove that $0$ is a lower bound of your sequence. In fact,$$(\forall n\in\mathbb{N}):n+1>n\implies\ln(n+1)>\ln n\implies\ln^2(n+1)>\ln^2n$$and therefore$$(\forall n\in\mathbb{N}):\ln^2(n+1)-\ln^2n>0.$$
A: You can write
$$
x_n=\ln\left(1+\frac{1}{n}\right)\ln n+\ln\left(1+\frac{1}{n}\right)\ln(n+1)
$$
Since
$$
\ln\left(1+\frac{1}{n}\right)\ln n<\ln\left(1+\frac{1}{n}\right)\ln(n+1)
$$
you just have to prove that
$$
y_n=\ln\left(1+\frac{1}{n}\right)\ln(n+1)
$$
is upper bounded. Now
$$
y_n=\ln\left(1+\frac{1}{n}\right)\ln n + \left(\ln\left(1+\frac{1}{n}\right)\right)^{\!2}
$$
The second summand is bounded because it has limit $0$. The first summand has limit zero as well, because
$$
\ln\left(1+\frac{1}{n}\right)\ln n<n\ln\left(1+\frac{1}{n}\right)
$$
and the sequence
$$
\left(1+\frac{1}{n}\right)^n
$$
is bounded. Actually,
$$
\lim_{n\to\infty}\ln\left(1+\frac{1}{n}\right)\ln n=0
$$
as you should be able to prove.
A: By mean value theorem for $f(x)=\ln^2x$ in $[n,n+1]$ there exists $n<\xi<n+1$ such that
$$\ln^2(n+1)-\ln^2(n)=2\dfrac{\ln\xi}{\xi}<2$$
because $\ln\xi<\xi$. Also $\ln$  is increasing then
$$\ln^2(n+1)-\ln^2(n)>0$$
A: Another concise option: $\ln(n+1)<\ln n+\frac{1}{n}$ so $0<\ln^2(n+1)-\ln^2 n<2\frac{\ln n}{n}+\frac{1}{n^2}<2\cdot 1+1=3$.
A: Lagrange's theorem provides a one-liner:
$$ \log^2(n+1)-\log^2(n) = \frac{d}{dx}\left.\log^2(x)\right|_{x=\xi\in(n,n+1)},\quad \frac{2\log\xi}{\xi}\to 0\text{ as }n\to +\infty. $$
