Proving using Bernoulli inequality

Question

I'm trying to prove this: $$0.99^n \le 1/2,\text{ for }n=100$$

I tried Bernoulli's inequality $(1-0.01)^n \geq 1-n\cdot0.01$ and it gave me LHS $\geq0$.

I also tried to do this: $((1-0.01)^n)^n \leq(1-1/2)^n$ and it gave me LHS $\geq-99$ and RHS $\geq49$.

Now I'm stuck.

Answer 1

Note that $$\left(1+\frac{1}{99}\right)^n \geq 1 +\frac{n}{99}$$ for every $n\geq 1$. In particular, $$\frac{1}{0.99^{100}}=\left(1+\frac{1}{99}\right)^{100}\geq 1+\frac{100}{99}>2\,.$$

Answer 2

Alternatively: $$0.99^{100}<\frac12 \iff 2\cdot 99^{100}<100^{100} \iff \\ 2\cdot 99^{100}<(1+99)^{100}=1+100\cdot 99+\cdots +100\cdot 99^{99}+99^{100} \iff \\ 2\cdot 99^{100}<1+9900+\cdots+(1+99)\cdot 99^{99}+99^{100}\iff \\ 2\cdot 99^{100}<1+9900+\cdots+99^{99}+2\cdot 99^{100}.$$