How to show that $x_n = \frac{n+1}{\log_2{(n+1)}}$ is unbounded without calculus?

Question

I want to prove the problem below without using calculus.

Let $n\in \mathbb N$ and: $$ x_n = \frac{n+1}{\log_2{(n+1)}} $$ Show that $x_n$ is unbounded.

I've started with comparing $x_{n-1}$ and $x_n$. Suppose $x_{n-1} > x_n$: $$ \begin{align} \frac{n}{\log_2n} &> \frac{n+1}{\log_2(n+1)} \iff \\ \iff n\log_2{(n+1)} &>(n+1)\log_2n \\ \log_2(n+1)^{n} &>\log_2n^{n+1} \\ \log_2(n+1)^{n} &> \log_2n + \log_2n^{n} \\ \log_2(n+1)^{n} - \log_2n^{n} &>\log_2n \\ \log_2\left(1+{1\over n}\right)^{n} &> \log_2n \end{align} $$

This is only true for $n=1$. Since $\left(1 + {1\over n}\right)^n$ is bounded by:

$$ \log_2{2} <\left(1 + {1\over n}\right)^n < \log_23 $$

and $\log_2n$ is unbounded we may conclude that $\{ \forall n > 1 : x_{n+1} > x_n\}$

But this only proves that the sequence is increasing starting from $n=2$. (It is actually $1+\sqrt5 \over 2$ if $n \in \mathbb R$). Since an increasing sequence is not necessarily unbounded the above doesn't solve what's in the problem statement.

Given $x_n$ is monotonically increasing starting from $n=2$ can we somehow prove it is unbounded?

Answer 1

Note that for $n=2^k-1$ and $k\geq 2$, $$\frac{n+1}{\log_2{(n+1)}}=\frac{2^k}{k}=\frac{(1+1)^k}{k}\geq \frac{\binom{k}{2}}{k}= \frac{k-1}{2}.$$

Answer 2

It is enough to show that $\frac{z}{\log_2 z}$ is unbounded as $z\to +\infty$, or that $\frac{2^w}{w}$ is unbounded as $w\to +\infty.$ On the other hand, assuming $w\geq 2$, $$ 2^w = \left(2^{w/2}\right)^2 \geq \left(1+\frac{w}{2}\right)^2 = 1+w+\frac{w^2}{4}$$ hence the last claim is trivial.

Answer 3

Let $M\ge 5$ then

$$\frac{n+1}{\log_2{(n+1)}} \ge M \iff n+1 \ge M\cdot \log_2{(n+1)}$$

and for $n+1\ge 2^M$

$$2^M\ge M\cdot \log_2{2^M}=M^2$$

wich is true.