If $n \ge 1$, does it follow that $\left(\frac{3}{2}\right)^n > n$

Question

It seems to me the answer is yes. Here's my reasoning:

For $n=1$, $\frac{3}{2} > 1$

and, $\frac{d}{dn}\left(\frac{\left(\frac{3}{2}\right)^n}{n}\right) > 0 $

Using the quotient rule:

$\dfrac{d}{dn}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} = \frac{\left(\frac{3}{2}\right)^n\left(\ln \frac{3}{2}\right)n - \left(\frac{3}{2}\right)^n }{n^2} > 0$ for $n \ge 1$

Am I correct? Is there a more standard way to show this?

Answer 1

We can observe that

$$\left(\frac{3}{2}\right)^n > n\iff 3^n >n2^n$$

and then proceed by induction.

Answer 2

Just check $f(x) = (\frac{3}{2})^x - x > 0$ for $x=1,2,3$.

Then for $x> 3$ you have $f'(x)= (\frac{3}{2})^x\ln \frac{3}{2} -1> 0$ anyway.

Answer 3

Here is an approach that is similar to that in your question, but perhaps a bit simpler.

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Since $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac{\log(x)}x=\frac{1-\log(x)}{x^2}\tag1 $$ The function $\frac{\log(x)}x$ is increasing for $x\lt e$ and decreasing for $x\gt e$ and thus has a maximum of $\frac1e$.

The inequality in the question is equivalent to $$ \frac{\log(n)}n\lt\log\left(\frac32\right)\tag2 $$ which is true since $\frac1e\lt\log\left(\frac32\right)$.

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Verification that $\boldsymbol{\frac1e\lt\log\left(\frac32\right)}$

If we can assume that $\frac52\lt e\lt\frac{11}4$, then $$ \frac1e\lt\frac25\tag3 $$ Furthermore, $$ \begin{align} e^2 &\lt\left(\frac{11}4\right)^2\\ &=\frac{121}{16}\\ &\lt\frac{243}{32}\\ &=\left(\frac32\right)^5\tag4 \end{align} $$ Combining $(3)$ and $(4)$ shows that $$ \frac1e\lt\frac25\lt\log\left(\frac32\right)\tag5 $$