Inflection points of $(1 - x^k) \mathbin{/} (1 - x^n)$ for $x$ strictly positive For $k$ and $n$ reals constants with $1 \leq k < n$, consider the function $f(x) = (1 - x^k) \mathbin{/} (1 - x^n)$, defined over the strictly positive reals ($0 < x$).
I am able to prove that $f(x)$ is decreasing and also that, if $k = 1$, $f(x)$ is convex.
But I cannot handle the case $1 < k$: I believe that $f(x)$ has exactly one inflection point (for $x$ strictly positive), but I am unable to prove it.
Remark: $f(x)$ is related to the gambler's ruin problem.
 A: After some more thinking, I can now answer my own question. The key of the proposed solution is Descartes' rule of signs. 
Let $h(x)$ be the second derivative of the function considered, that is $h(x) = (d/dx)^2\; (1 - x^k) \mathbin{/} (1 - x^n)$. After some computations, $h(x)$ can be expressed as a generalised (the exponents are not necessarily integers) rational function of $x$:$$
    x^{k - 2} \cdot \frac{
      + \;((n - k)^2 + (n - k)) \cdot x^{2 \cdot n} \\
      - \;n \cdot (n + 1) \cdot x^{2 \cdot n - k} \\
      + \;(n \cdot (n + 1) + 2 \cdot (k - 1) \cdot (n - k)) \cdot x^n \\
      - \;n \cdot (n - 1) \cdot x^{n - k} \\
      + \;(k - 1) \cdot k
    }{(x^n - 1)^3}
$$
The goal is to show that $h(x)$ has, for $1 < k$ and $0 < x$ (both conditions hold implicitly from now on), exactly one sign change.
The denominator of $h(x)$ has exactly one root, at $x = 1$, of multiplicity 3. It can be shown, with L'Hospital's rule, that, up to continuous continuation, $h(1) = k \cdot (n - k) \cdot (n - 2 \cdot k + 3) \mathbin{/} (6 \cdot n)$. In particular, the numerator of $h(x)$ has a root of multiplicity at least 3 at $x = 1$.
The ordered coefficients of the numerator of $h(x)$ (a generalised polynomial in $x$) have 4 sign changes, and thus, according to Descartes' rule of signs, 0, 2 or 4 positive roots, counted with multiplicities. Hence, the numerator of $h(x)$ has exactly 4 positive roots, counted with multiplicities: one root of multiplicity 3 at $x = 1$, and one more root of multiplicity 1 at $x = r$ for some strictly positive $r$ (if $n - 2 \cdot k - 3 = 0$, then $r = 1$ as well).
Since the numerator of $h(x)$ has one root of multiplicity 3 (or 4) at $x = 1$, thus canceling the root of multiplicity 3 at $x = 1$ of its denominator, $h(x)$ remains with exactly one positive root, at $x = r$, of multiplicity 1. That is, $h(x)$ has exactly one sign change (for $x$ with $0 < x$).
Remark: For $x$ strictly positive and big enough, $h(x)$ is positive, that is $(1 - x^k) \mathbin{/} (1 - x^n)$ is convex.
Remark: If $k = 1$, then the ordered coefficients of the numerator of $h(x)$ have only 3 sign changes, and thus, by the same device as above, $h(x)$ has no sign change, that is $(1 - x^k) \mathbin{/} (1 - x^n)$ is convex (for $x$ with $0 < x$).
Remark: The same method can be used to show that $(1 - x^k) \mathbin{/} (1 - x^n)$ is decreasing in $x$ (for $x$ with $0 < x$).
