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Ermakoff's test states that for a nonincreasing function $f(x)$, $\sum f(n)$ converges if

$$\limsup_{x \to \infty} \frac{e^x f(e^x)}{f(x)} < 1$$

and diverges if

$$\liminf_{x \to \infty} \frac{e^x f(e^x)}{f(x)} > 1.$$

This test seems to be very powerful in that any example I try gives a limit of $0$ or $\infty$.

Question: Is there an example where Ermakoff's test is inconclusive, i.e. is there an $f(x)$ that gives a limit of 1? More generally, is there an $f(x)$ that gives a finite positive limit?

My approach was to toe the line between slowly diverging summands like $\frac{1}{n}, \frac{1}{n \log n}, \frac{1}{n \log n \log\log n}, \ldots$ I tried throwing in special functions that grow differently than a power of a nested logarithm, such as $li(x)$ and $W(x)$. This didn't get me very far as I ended up wandering aimlessly through examples.

By taking the logarithm of the limit and performing substitutions, it's equivalent to look for a nondecreasing function $g(x)$ where $\lim_{x \to \infty} (x - g(x) + g(\log x)) = 0$. This form makes working with series at $x = \infty$ a bit easier.

The test can be generalized by looking at $\phi'(x) f(\phi(x))/f(x)$ for any increasing $\phi(x) > x$. Letting $\phi(x) = e^x$ gives Ermakoff's test and letting $\phi(x) = x + 1$ gives d'Alembert's ratio test.

Bonus question: Is there a way to find an inconclusive example for any $\phi(x)$? I imagine the faster $\phi(x)$ grows, the harder it becomes.

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  • $\begingroup$ My script, nfnitseriestst.m (written in MATLAB and on the File Exchange) shows through testing that Ermakoff's test is indeed good at logarithm's, and is a powerful convergence test. I have not found a series where the limit is equal to 1. The limit is always zero or Inf, or doesn't exist. $\endgroup$ Commented Nov 9, 2022 at 0:21

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We can find a counterexample by treating the limit in question as a functional equation (a stronger condition), i.e. solve the functional equation

$$ f(e^x) = f(x)e^{-x} \implies f(x) = f(\log x)/x. $$

Choosing the initial values $f(x) = 1$ for $0 \leq x < 1$ yields a continuous solution through forward propogation:

$$ f(x) = \begin{cases} 1 & 0 \leq x < 1 \\ \frac{1}{x} & 1 \leq x < e \\ \frac{1}{x \log x} & e \leq x < e^e \\ \frac{1}{x \log x \log \log x} & e^e \leq x < e^{e^e} \\ \frac{1}{x \log x \log \log x \log\log\log x} & e^{e^e} \leq x < e^{e^{e^e}} \\ \vdots & \vdots \end{cases} $$

This can be written more concisely in terms of the iterated logarithm (log star).

Define $\log^*\!x$ as the number of times $\log$ needs to be applied to $x$ until it's less than 1 and $\log_k x$ to be $k$ iterations of $\log$, i.e.

$$ \log^*\!x = \begin{cases} 0 & x < 1 \\ 1 + \log^*(\log x) & x \geq 1 \end{cases} \quad \text{and} \quad \log_k x = \underset{k}{\underbrace{\log \log \cdots \log }}\; x.$$

Then $f(x)$ can be written as

$$ f(x) = \prod_{k = 0}^{\log^*\!x-1} \frac{1}{\log_k x}. $$

By construction Ermakoff's test is inconclusive for $f(x)$, and the choices $\phi(x) = e^{e^x}$, $\phi(x) = e^{e^{e^x}}$, $\phi(x) = e^{e^{e^{e^x}}}$, $\ldots$ are all inconclusive too. One must take $\phi(x) = \text{tet}(x)$ (tetration), or anything that grows faster, to show this sum diverges.

Additionally, the integral test easily shows the divergence of the sum.

This approach can apply to any $\phi(x) > x$ where $\phi''(x) > 0$ and there is an $a > 0$ such that $\phi'(\phi^{-1}(a)) = 1$.

Let $\psi(x) = \phi^{-1}(x)$ and choose $a \geq 0$ such that $\phi'(\psi(a)) = 1$.

We can construct a continuous function satisfying

$$ f(\phi(x)) = f(x)/\phi'(x) \quad \text{and} \quad f(x) = 1 \;\; \text{for} \;\; 0 \leq x < a $$

through the same approach as above:

$$ f(x) = \begin{cases} 1 & 0 \leq x < a \\ \frac{1}{\phi'(\psi(x))} & a \leq x < \phi(a) \\ \frac{1}{\phi'(\psi(x))\phi'(\psi(\psi(x))} & \phi(a) \leq x < \phi(\phi(a)) \\ \frac{1}{\phi'(\psi(x))\phi'(\psi(\psi(x))\phi'(\psi(\psi(\psi(x)))} & \phi(\phi(a)) \leq x < \phi(\phi(\phi(a))) \\ \vdots & \vdots \end{cases} $$

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