3

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?

user144945
  • 33
  • The generalization fails gecause $x^2+x^3$ is not strictl yincreasing. – Hagen von Eitzen Apr 22 '14 at 18:03
  • Right, thanks. Is there a more mild generalization that is suitable for induction in the same way? k-times differentiation of $\sum_{i=0}^n x^i$ yields $\sum_{i=0}^{n-k} \frac{(i+k)!}{i!}x^i$. Perhaps it holds for all polynomials of that form for all $k$. – user144945 Apr 22 '14 at 18:11

1 Answers1

1

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.

André Nicolas
  • 507,029