# Prove $1.01^{1000} > 1000$ without using calculator

Prove $$1.01^{1000} > 1000$$ without using calculator.

With WolframAlpha $$1.01^{1000} \approx 20959$$, but can this be proved without calculator?

By Bernoulli's inequality, $$\left(1+\frac1{100}\right)^{100}\ge 1+\frac1{100}\cdot 100,$$ so $$(1.01)^{1000}=((1.01)^{100})^{10}\ge 2^{10}.$$

If this is true $$1.01^{1000} > 1000$$ then $$1000\log_{10} \left(1+\frac{1}{100}\right) > 3\implies \frac{1000}{\log(10)}\log \left(1+\frac{1}{100}\right) > 3$$ and $$\log \left(1+\frac{1}{100}\right)\sim \frac{1}{100}$$ So $$\frac{1000}{\log(10)}\log \left(1+\frac{1}{100}\right)\sim \frac{10}{\log(10)}=\frac{10}{2.3026}>\frac{10}{2.5}=4$$

• But $\log \left(1+\frac{1}{100}\right)< \frac{1}{100}$, and you are using it for an estimate in the other direction. The approximation makes it plausible that the desired inequality holds, but without an error estimate it is not a strict proof. Feb 4, 2020 at 8:15

Using Hagen von Eitzen's idea, we can prove the following extension:

Consider two positive integers $$m$$ and $$n$$ such that $$m \geq n+2$$. Then:

$$1.\underbrace{00\ldots 0}_\text{n zeros}1^{10^m} > 1000^{10^{m-n-2}}$$

Indeed, using Bernoulli

$$1.\underbrace{00\ldots 0}_\text{n zeros}1^{10^m} = \left[\left(1+\frac {1}{10^{n+1}}\right)^{10^{n+1}}\right]^{10^{m-n-1}}\ \ge\ \left(1+10^{n+1}\cdot\frac {1}{10^{n+1}}\right)^{10^{m-n-1}}$$ $$=2^{10^{m-n-1}}=1024^{10^{m-n-2}}>1000^{10^{m-n-2}}$$

The OP is the case of $$m=3$$ and $$n = 1$$.