bernoulli inequality

Bernoulli inequality

For $x\ge -1$ and $n\in \mathbb{N}$:

$$(1+x{)}^n\ge 1+nx$$

Equality holds iff $n=1$ or $x=0$.

It provides a linear lower bound on $(1+x{)}^n$.

Proof (induction)

Base case ($n=1$): $(1+x{)}^1=1+x$.

Inductive step: Assume $(1+x{)}^n\ge 1+nx$. Then:

$$(1+x{)}^{n+1}=(1+x{)}^n(1+x)\ge (1+nx)(1+x)=1+(n+1)x+n{x}^2\ge 1+(n+1)x$$

since $n{x}^2\ge 0$.

Application:

Proof that $z^n\to 0$ for $|z|<1$

For $z=0$ it’s trivial. For $z\ne 0$, write $|z|=\frac{1}{1+x}$ with $x>0$.
By the bernoulli inequality, $(1+x{)}^n\ge 1+nx$, so:

$$|z^n|=\frac{1}{(1+x{)}^n}\le \frac{1}{1+nx}\le \frac{1}{nx}=\frac{1}{x}\cdot \frac{1}{n}$$

Since $\frac{1}{n}\to 0$, we have $z^n\to 0$.

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Quadratic bound

For $x\ge 0$, the binomial expansion gives a tighter bound:

$$(1+x{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k\ge 1+nx+\frac{n(n-1)}{2}{x}^2$$

Used when the linear term cancels or is too weak (e.g., proving $\sqrt[n]{n}\to 1$).