Theorem (Bernoulli's Inequality) Let \(\alpha\) be a positive real number and \(\delta\geq -1\). If \(0< \alpha\leq 1\), the \[ (1+\delta)^\alpha \leq 1+\alpha\delta, \] and if \(\alpha\geq 1\), then \[ (1+\delta)^\alpha\geq 1+\alpha\delta. \]
Proof. Assume \(0< \alpha\leq 1\). Let \(f(x)=x^\alpha\). By the mean value theorem, \[ f(1+\delta)=f(1)+\alpha\delta c^{\alpha-1} \] for some \(c\) between \(1\) and \(1+\delta\). If \(\delta>0\), then \(c>1\). Since \(0< \alpha\leq 1\), it follows that \(c^{\alpha-1}\leq 1\). Hence, \(\delta c^{\alpha-1}\leq \delta\). On the other hand, if \(-1\leq \delta\leq 0\), then \(c^{\alpha-1}\geq 1\) and again \(\delta c^{\alpha-1}\leq\delta\). Thus, \[ (1+\delta)^\alpha=f(1+\delta)=f(1)+\alpha\delta c^{\alpha-1}\leq 1+\alpha \delta. \] Now we will prove the case \(\alpha\geq 1\). Assume \(\alpha\geq 1\) and let \(f(x)=x^\alpha\). Using the MVT, again, \[ f(1+\delta)=f(1)+\alpha\delta c^{\alpha-1} \] for some \(c\) between \(1\) and \(1+\delta\). If \(\delta>0\), then \(c>1\). Since \(\alpha\geq 1\), it follows that \(c^{\alpha-1}\geq 1\). Hence, \(\delta c^{\alpha-1}\geq \delta\). On the other hand, if \(-1\leq \delta\leq 0\), then \(c^{\alpha-1}\leq 1\) and again \(\delta c^{\alpha-1}\geq\delta\). Thus, \[ (1+\delta)^\alpha=f(1+\delta)=f(1)+\alpha\delta c^{\alpha-1}\geq 1+\alpha \delta, \] and we have the conclusion.
Exercise 1. Prove that \[ a_n=\left(1+\frac{1}{n}\right)^n \] is a monotone increasing sequence.
Solution Since by Bernoulli's inequality, we have \[ \left(1+\frac{1}{n}\right)^{\frac{n}{n+1}}\leq \left(1+\frac{1}{n+1}\right). \]