Is it possible that $nt + (1-t^n) - n > 0$ for $n > 1$ and $t \in [0,1]$?

Question

Is it possible that $nt + (1-t^n) - n > 0$ for $n > 1$ and $t \in [0,1]$? I'm assuming here that $n \in \mathbb{N}$.

We know that

$$ nt + (1-t^n) - n = \underbrace{(nt - n)}_{\le 0} + \underbrace{(1 - t^n)}_{ \in [0,1]} $$

However, I'm not sure how to show that this quantity is less than or equal to $0$.

Answer 1

Let $f(t)=nt+(1-t^n)-n$. We show that $f(t)\le 0$ in our interval, meaning that it is not possible that $f(t)\gt 0$ in the interval $[1,0]$.

Note that $f(0)=-(n-1)\lt 0$. Also, $f'(t)=n-nt^{n-1} \gt 0$ if $0\lt t\lt 1$ and $n \gt 1$, so $f(t)$ is steadily increasing in our interval. Since $f(1)=0$, $f(t)$ cannot be positive anywhere in the interval $[0,1]$.

Remark: We are probably intended to assume that $n$ is an integer. However, the above argument works for any real number $n\gt 1$.

Answer 2

Note that $1-t^n=(1-t)(1+t+\dots t^{n-1})\le (1-t)\cdot n$ when $|t|\le 1$and you are pretty much home when you join the dots and deal with the potential case(s) of equality.

The factorisation of $1-t^n$ is one you should remember, as it will come in handy again.