Question

For which natural numbers "n" is irrelevance?

x - real number
x > -1

`(1+x)^n > 1+nx`

have to prove

Answer 1

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

Explanation:

Let `f(x)=(1+x)^n` and `g(x)=1+nx` for `n>1`. Note that both `f` and `g` are continuous and differentiable for all `x` and that `f(0)=g(0)=1`.

Next, we compute `f'(x)=n(1+x)^(n-1)` and `g'(x)=n`. Since `n>1`, we have `n-1>0`.

Therefore, for `x>0`, `f'(x)>n*1^(n-1)=n=g'(x)`.

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(x) > g(x)` for `x>0` and for `-1 < x < 0`. In other words, `(1+x)^(n) > 1+nx` for `x>0` and for `-1 < x < 0` (they are equal when `x=0`).