The Most Powerful Criterion for Convergence of Positive-Term Series?
When studying positive-term series, our textbooks on mathematical analysis provide various convergence tests: the integral test, the comparison test, and the root test and ratio test derived from it, culminating in the rather refined "Raabe's test." These methods differ in precision (strength), and Raabe's test is generally considered to have the widest range of applicability. But in my view, the comparison test based on p-series can already handle every problem — it is actually the strongest method.
The p-series is the familiar
$$\sum_{n=1}^{\infty} \frac{1}{n^p}$$
Using the integral test, one can show that this series converges when p>1, and diverges otherwise. Although I can't prove it, I believe the following statement holds:
If a positive-term series $\sum_{n=1}^{\infty} a_n$ converges, then one can always find a constant A and a constant p greater than 1 such that every term satisfies $a_n < \frac{A}{n^p}$.
Since the convergence and divergence of the p-series is itself established via the integral test, the integral test is, in the most fundamental sense, the underlying method. However, it's not always convenient to apply directly, so what we actually discuss here is the "comparison test based on p-series."
Let's use an example to illustrate how to find A and p.
$$\sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!} \frac{1}{2n+1}$$
where !! denotes the double factorial, i.e., $n!! =n(n-2)(n-4)...$. The series above converges, but its conditions are quite delicate — the usual tests don't apply here. The textbook approach is to use Raabe's test, which indicates that this is a series with relatively slow convergence. Below, we prove its convergence using the comparison test instead.
Let
$$\frac{(2n-1)!!}{(2n)!!} \frac{1}{2n+1} < \frac{A}{n^p}$$
The main idea is mathematical induction: for n+1, we have:
$$\frac{(2n+1)!!}{(2n+2)!!} \frac{1}{2n+3} < \frac{A}{(n+1)^p}$$
Dividing the two sides, we find that a sufficient condition for the former to imply the latter is:
$$(\frac{2n+1}{2n+2})(\frac{2n+1}{2n+3}) < (\frac{n}{n+1})^p$$
We can always choose a sufficiently small p greater than 1 so that the above inequality holds for all n; let's first estimate what p should be. Rewrite it as:
$$(\frac{2+\frac{1}{n} }{2+\frac{2}{n}})(\frac{2+\frac{1}{n}}{2+\frac{3}{n}}) < (\frac{1}{1+\frac{1}{n}})^p$$
Expanding both sides in terms of $\frac{1}{n}$ (note that at this stage we are only estimating, not yet proving):
$$1-\frac{1.5}{n} < 1-\frac{p}{n}$$
A preliminary estimate suggests p=1.5 should suffice; to be safe, one could also try p=1.4, 1.3, etc. It turns out that with p=1.5, the inequality
$$(\frac{2n+1}{2n+2})(\frac{2n+1}{2n+3}) < (\frac{n}{n+1})^p$$
already holds for all n (I won't write out the proof in detail — it's fairly simple). How should we choose A? In fact, once p is fixed, we only need to pick A large enough that the first term of $\frac{A}{n^p}$ exceeds the first term of the original series — here that's simple, and A=1 works. With that, the proof is complete: by constructing
$$\frac{(2n+1)!!}{(2n+2)!!} \frac{1}{2n+3} < \frac{1}{(n+1)^{1.5}}$$
we've finished the proof.
On the other hand, proving divergence is even easier, since we don't need to choose a suitable p at all (just take p=1) — we only need to pick an appropriate A!
Conclusion
Of course, the claim in this post that the p-series comparison test is the strongest is merely a conjecture. Whether there exists some positive-term series that cannot be handled by this method, I honestly don't know. But based on my own experience solving problems, I have yet to encounter an example that this method fails to handle. If a reader can find such a counterexample, that would be even better.
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.