# For which natural numbers "n" is irrelevance?

## x - real number x > -1 ${\left(1 + x\right)}^{n} > 1 + n x$ have to prove

Feb 25, 2018

$n > 1$ for this to be true. You can use the "Racetrack Principle " or the Mean Value Theorem to prove it.

#### Explanation:

Let $f \left(x\right) = {\left(1 + x\right)}^{n}$ and $g \left(x\right) = 1 + n x$ for $n > 1$. Note that both $f$ and $g$ are continuous and differentiable for all $x$ and that $f \left(0\right) = g \left(0\right) = 1$.

Next, we compute $f ' \left(x\right) = n {\left(1 + x\right)}^{n - 1}$ and $g ' \left(x\right) = n$. Since $n > 1$, we have $n - 1 > 0$.

Therefore, for $x > 0$, $f ' \left(x\right) > n \cdot {1}^{n - 1} = n = g ' \left(x\right)$.

On the other hand, for $- 1 < x < 0$ we have $0 < 1 + x < 1$ and f'(x) < n*1^(n-1)=n=g'(x).

Putting these facts together and applying the Racetrack Principal allows us to say that $f \left(x\right) > g \left(x\right)$ for $x > 0$ and for $- 1 < x < 0$. In other words, ${\left(1 + x\right)}^{n} > 1 + n x$ for $x > 0$ and for $- 1 < x < 0$ (they are equal when $x = 0$).