How do you solve this inequality? $\frac{1}{(n+1)4^{n+1}} < .001 $

Question

$$\frac{1}{(n+1)4^{n+1}} < .001 $$

becomes

$$ 1000 < (n+1)4^{n+1}$$

Where do you go from here? Am I supposed to plug in a table of values for n?

n=1: 1000 < 32

n=2: 1000 < 192

n=3: 1000 < 1024

So, n=3 satisfies the inequality.

But, I've been told the answer is $n>3$

Answer 1

Assuming that we're solving the inequality over $\mathbb{N}$, we may use induction. First, one can already sense that there exists some $N$ such that the inequality, $$1000\lt(n+1)4^{n+1},\tag1$$ will be true for all $n\geqslant N$. A reasonable guess for such $N$ is, as you found, $N=3$. Now suppose that $(1)$ is true. Then $$ \begin{align} 1000\lt(n+1)4^{n+1} & \implies 1000\lt(n+2)4^{n+1}\lt (n+2)4^{n+1}\cdot 4 \\ &\implies 1000\lt(n+2) 4^{n+2}. \end{align}$$ Hence by induction, $(1)$ holds for all $n\geqslant3$.

Answer 2

To solve for $n$, I must first start with $1000=(n+1)4^{n+1}$

The solution for $n$ may be found in terms of the Lambert W function (or product log), and it is:

$$n=\frac{W(1000\ln(4))}{\ln(4)}-1$$

Where $W(x)$ is the Lambert W function.

In other words, to solve the inequality, it is simply $n>\frac{W(1000\ln(4))}{\ln(4)}-1\approx2.985509972\approx3$

So there you have it.