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Bernoulli Inequality

(1+x)^n > 1+nx,
\end{displaymath} (1)

where $x \in \Bbb{R} > -1 \not= 0$, $n \in \Bbb{Z} > 1$. This inequality can be proven by taking a Maclaurin Series of $(1+x)^n$,
(1+x)^n=1+nx+{\textstyle{1\over 2}}n(n-1)x^2+{\textstyle{1\over 6}} n(n-1)(n-2)x^3+\ldots.
\end{displaymath} (2)

Since the series terminates after a finite number of terms for Integral $n$, the Bernoulli inequality for $x>0$ is obtained by truncating after the first-order term. When $-1<x<0$, slightly more finesse is needed. In this case, let $y=\vert x\vert=-x>0$ so that $0<y<1$, and take
(1-y)^n=1-ny+{\textstyle{1\over 2}}n(n-1)y^2-{\textstyle{1\over 6}} n(n-1)(n-2)y^3+\ldots.
\end{displaymath} (3)

Since each Power of $y$ multiplies by a number $<1$ and since the Absolute Value of the Coefficient of each subsequent term is smaller than the last, it follows that the sum of the third order and subsequent terms is a Positive number. Therefore,
\end{displaymath} (4)

(1+x)^n>1+nx,\qquad{\rm for\ } -1<x<0,
\end{displaymath} (5)

completing the proof of the Inequality over all ranges of parameters.

© 1996-9 Eric W. Weisstein