Upper bound for $\prod\limits_{k = 1}^{n}\left( 1 - x^{k+1}\right)$

Question

For a real $x \in [0, 1)$, does anyone know a non-trivial upper bound for $ \prod_{k = 1}^{n}\left( 1 - x^{k+1}\right) $? Like the fellow here, I am sure I've something like this somewhere... Notice that in my case an upper bound suffices.

Answer 1

Write $f(x)$ for your function. I'll be detailed here since this is a nice pedagogical example. Since all the terms are positive an equivalent problem is to upper bound the logarithm

$$\log f(x) = \sum_{k=1}^n \log (1 - x^{k+1}).$$

Any upper bound on the function $\log (1 - r)$ on the interval $r \in [0, 1)$ then provides a bound on this sum. Let's remind ourselves what $\log (1 - r)$ looks like; here is a plot from WolframAlpha:

In particular it's decreasing and concave. $0$ is a trivial upper bound, which gives the trivial upper bound $1$ on the product as mentioned in the comments. A more interesting upper bound is the linear Taylor approximation at the origin, which is $1 - r$. Here's a plot of the two of them together:

This gives us the famous inequality $\log (1 - r) \le -r$, which may be more familiar in exponential form $1 - r \le e^{-r}$. Applying this inequality to each term of the sum above gives a sum of geometric series

$$\log f(x) \le \sum_{k=1}^n -x^{k+1} = -x^2 \left( \frac{1 - x^n}{1 - x} \right)$$

and exponentiating gives

$$\boxed{ f(x) \le \exp \left( -x^2 \left( \frac{1 - x^n}{1 - x} \right) \right) }.$$

This bound is designed to be particularly good for small $x$; it becomes less and less accurate for larger $x$ since the same is true of the Taylor approximation we used, and note that at $x = 1$ it gives $f(1) \le \exp (-n)$ whereas in fact $\lim_{x \nearrow 1} f(x) = 0$.

Fortunately, since $\ln (1 - r)$ is concave it lies below all of its tangent lines, so depending on what values of $x$ you're interested in you can replace the bound we used above by other linear bounds. For example, the linear Taylor approximation at $r = \frac{1}{2}$ gives

$$\ln (1 - r) \le -2r - \ln 2 + 1.$$

Here's what this one looks like:

Of course, as other commenters have said, it's hard to know what kind of bounding is appropriate here if we don't know what you're planning on using them for; for some purposes the bound $f(x) \le 1$ might suffice.