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?
Best Answer
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$$(\forall n\in\mathbb{N}):\ln^2(n+1)-\ln^2n>0.$$