Verify the proof that $x_n = \ln^2(n+1) - \ln^2n$ is a bounded sequence.

Question

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?

Answer 1

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 $$\ln^2(n+1)-\ln^2(n)>0$$

Answer 2

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

Answer 3

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.

Answer 4

Another concise option: $\ln(n+1)<\ln n+\frac{1}{n}$ so $0<\ln^2(n+1)-\ln^2 n<2\frac{\ln n}{n}+\frac{1}{n^2}<2\cdot 1+1=3$.

Answer 5

Lagrange's theorem provides a one-liner:

$$ \log^2(n+1)-\log^2(n) = \frac{d}{dx}\left.\log^2(x)\right|_{x=\xi\in(n,n+1)},\quad \frac{2\log\xi}{\xi}\to 0\text{ as }n\to +\infty. $$