A little unknown test for convergence of series with positive terms

Question

While helping a highschooler studying for her Calculus class, she showed me a convergence test for series which is kind of a hybrid between the ratio and root test, and of which I was not aware:

Theorem: Suppose $a_n>0$ for all $n$.

1. If $\limsup_n\Big(\frac{a_{n+1}}{a_n}\Big)^n<\frac1e$, then the series $\sum_na_n$ converges.
2. If there is $N$ such that for all $n\geq N$, $\Big(\frac{a_{n+1}}{a_n}\Big)^n\geq\frac1e$, then the series $\sum_na_n$ diverges.

I put this test to the test with simple examples:

• $\sum_n\frac1n$: $\Big(\frac{n+1}{n}\Big)^n=\big(1+\tfrac1n\big)^n\nearrow e$ and so, with $a_n=\frac1n$, $\Big(\frac{a_{n+1}}{a_n}\Big)^n\geq\frac1e$ which means that $\sum_n\frac1n$ diverges, as it should.

• $\sum_n\frac1{n^p}$: $\Big(\frac{n}{n+1}\Big)^{pn}=\frac{1}{\big(1+\frac1n\big)^{np}}\xrightarrow{n\rightarrow\infty}\frac{1}{e^p}$. For $p>1$ we get convergence and for $p<1$ divergence, as it should

• $\sum_n\frac{(a)_n(b_n)}{(c)_nn!}$, where $(z)_0:=1$ and $(z)_n=z\cdot\ldots\cdot(z+n-1)$ for $n\geq1$, $z\in \mathbb{C}$. For simplicity, assume that $a,b,c\in\mathbb{R}\setminus\mathbb{Z}_-$. Let $u_n=\Big|\frac{(a)_n(b_n)}{(c)_nn!}\Big|$. Then, for all $n$ large enough $$\Big(\frac{u_{n+1}}{u_n}\big)^n=\Big(\frac{(a+n)(b+n)}{(n+1)(n+c)}\Big)^n= \Big(1+\frac{a-1}{n+1}\Big)^n\Big(1+\frac{b-c}{n+1}\Big)^n\xrightarrow{n\rightarrow\infty}e^{a+b-c-1}$$ The series $\sum_nu_n$ converges if $a+b-c<0$ and diverges if $a+b-c>0$. This can also be obtained by Raabe's test for example.

In fact, the Theorem above seems to be at the same level as Raabe's test in the de Morgan hierarchy.

Question: Does anybody know of the provenance of the Theorem above amd/or a reference?

Thanks!

---

Edit: Just to present a short proof of the theorem above:

(1) Let $p>1$ be such that $\limsup_n\Big(\frac{a_{n+1}}{a_n}\Big)^n<\frac{1}{e^p}<\frac{1}{e}$ Then, there is $N$ such that for all $n\geq N$ $$a_{n+1}<e^{-p/n}a_n$$ Let $H_n:=\sum^n_{k=1}\frac1k$ the $n$-th sum of the harmonic series. It follows that $$a_{n+1}<\exp(-pH_n)e^{H_N}a_N<c_Nn^{-p},\qquad n\geq N$$ for some constant $C_N>0$.

(2) The assumption here implies that for all $n\geq N$ $$a_{n+1}\geq e^{-1/n}a_n$$ and so, $$a_{n+1}\geq e^{-\big(\tfrac{1}{n}+\ldots+\tfrac{1}{N}\big)}a_N=e^{-H_n}e^{H_N}a_N\geq \frac{C_N}{n}$$ for some constant $c_N>0$. Hence $\sum_na_n$ diverges.

Answer 1

(edited from comments)

As far as I know, this result was first proved relatively recently (for a result such as this):

Orrin Frink, A ratio test, Bulletin of the American Mathematical Society 54 #10 (October 1948), p. 953.

I thought Frink's paper was relatively well known, but google searches (regular & book & scholar) seem to show otherwise.

I came across Frink's paper roughly 41-42 years ago while browsing through library journal volumes of Bull. AMS. The paper caught my attention at the time (and it was one of the first papers I ever made a photocopy of) by being one of only a handful of papers anywhere in Bull. AMS that, at the time, I had the background to follow everything.

Regarding Martin R's comment about the relation of Frink's test to Raabe's test, see Marceli Stark, On a ratio test of Frink, Colloquium Mathematicum 2 #1 (1949), pp. 46-47. Note: In several web sites the title of Stark's paper is incorrectly given as "On the ratio test of Frink". Also, the footnote in Stark's paper incorrectly gives Vol. 55 for Frink's paper.