polynomials - monotonicity of $sum_0^n x^i$ for odd $n$

I am trying to prove the strict monotonicity of $\sum_{i=0}^n x^i$ for odd $n$. This is not homework; just something I have noticed to appear true, and thus my brain bugs me until I have a proof.

I have tried a direct approach but am coming up empty.

Another approach I have tried is to prove that $\sum_{i=0}^n a_ix^i$ for $0 < a_0 \le a_1 \le \ldots \le a_n$ is strictly increasing for $n$ odd and $\ge 0$ with equality holding at at most one point for $n$ even. I am not sure whether this claim is true (edit: it is false but a more mild generalization may hold - see comments) but it is sufficient to show the original claim and it appears to lend itself to a proof by induction alternating between the even and odd cases. Specifically, the base cases of $n\in\{0, 1\}$ are obvious. If the claim holds for some even $n$, it holds for $n+1$ because its derivative is of the form in the claim and must be increasing at all but one point. Also, if the claim holds for some odd $n$, the $n+1$ case must have an increasing derivative; furthermore, the derivative goes from arbitrarily small for small $x$ to arbitrarily large for large $x$, so the $n+1$ case must be U shaped and thus must have a single minimum. If we could show that this minimum $\ge 0$, we would be done, but I am not sure whether it is.

I don't know whether to continue the inductive approach or whether to try something else. Does anyone have any ideas, advice, or solutions?

Answer

It is clear that our function is strictly increasing on the interval $[0,\infty)$. To show it is increasing on the negatives, let
$$f(x)=1+x+x^2+\cdots+x^{2m+1}=\frac{x^{2m+2}-1}{x-1}.$$
Differentiate. We get
$$\frac{1-(2m+2)x^{2m+1}+(2m+1)x^{2m+2}}{(x-1)^2},$$
which is positive if $x$ is negative.