Prove $1.01^{1000} > 1000$ without using calculator.
With WolframAlpha $1.01^{1000} \approx 20959$, but can this be proved without calculator?
Prove $1.01^{1000} > 1000$ without using calculator
inequality
Related Solutions
$$ 1.05^{50} = \left(1+\frac{5}{100}\right)^{50} = \sqrt{\left(1+\frac{5}{100}\right)^{100}} < \sqrt{e^5} = \sqrt{e}^5 < 2^5 = 32 $$
Edit: for you second inequality, $2^{1000}<10^{302}$ is equivalent to $(2^{10})^{100}<10^2(10^3)^{100}$, which is equivalent to $$ \left(\frac{2^{10}}{10^3}\right)^{100} = \left(1+\frac{24}{1000}\right)^{100} < 100. $$ From $a\log(1+t)\leq at$ for $a>0$ we deduce $(1+t)^a\leq e^{at}$. Therefore $$ \left(1+\frac{24}{1000}\right)^{100} < e^{\frac{24}{1000}\cdot100} <e^{5/2} < 32 < 100. $$
Addendum. What is trickier, is proving that $2^{1000}>10^{301}$. Can you do that?
Here is how I go about it. Maybe someone else can find a simpler derivation.
Define $$ \exp_n(x) = \sum_{k=0}^n \frac{x^k}{k!} < \exp(x) . $$
We want $\left(\frac{2^{10}}{10^3}\right)^{100}>10$. From $(1-t)^a\leq e^{-at}$ you deduce $\left(\frac1{1-t}\right)^a\geq e^{at}$. So $$ \begin{split} \left(\frac{2^{10}}{10^3}\right)^{100} &= \left(\frac{1024}{1000}\right)^{100} = \left(\frac{1}{1-\frac{24}{1024}}\right)^{100} \geq e^{100\cdot\frac{24}{1024}} = e^{75/32} \\ &= e^{2+11/32} > e^{2+11/33} = \exp(2)\exp(1/3) > \exp_5(2)\exp_2(1/3) \\ &= \left(1+2+2+\frac43+\frac23+\frac4{15}\right)\left(1+\frac13+\frac1{18}\right) \\ &= \frac{109}{15} \cdot \frac{25}{18} = \frac{545}{54} > 10. \end{split} $$
Let's call $\begin{cases}R=\ln\left(\dfrac{2^{123}}{5^{53}}\right)=123\ln(2)-53\ln(5)\\\\a=\ln(1.024)\end{cases}$
We expand the logarithm of $2$ now:
$10\ln(2) = \ln(1024) = \ln(1.024)+3\ln(10) = a + 3\ln(2)+3\ln(5)$
$$a = 7\ln(2)-3\ln(5)$$
And we use it to reduce $R$:
$R=119\ln(2)+4\ln(2)-51\ln(5)-2\ln(5)=17a+4\ln(2)-2\ln(5)=17a-\ln(\frac{25}{16})$
We will now use the inequality $$\ln(1+x)\le x\le\frac 12\ln\left(\dfrac{1+x}{1-x}\right)$$
Note that $\frac{1+x}{1-x}=y\iff x=\frac{y-1}{y+1}$
So we have $\begin{cases}17a\le \frac{17\times 24}{1000}=\frac{51}{125}\\\\ \ln(\frac{25}{16})\ge 2\times\frac{25/16-1}{25/16+1}=\frac{18}{41}\end{cases}$
This allows us to conclude that $R\le \dfrac{51}{125}-\dfrac{18}{41}=-\dfrac{159}{5125}<0$
Numerical verification:
$R \approx -0.0431 < -\frac{159}{5125}\approx -0.03102$
Since $R<0\iff e^R<1$ we conclude that $2^{123}<5^{53}$
Best Answer
By Bernoulli's inequality, $$\left(1+\frac1{100}\right)^{100}\ge 1+\frac1{100}\cdot 100, $$ so $$(1.01)^{1000}=((1.01)^{100})^{10}\ge 2^{10}. $$