The Most Powerful Convergence Test for Series with Positive Terms?

When studying series with positive terms, our mathematical analysis textbooks provide various convergence tests, such as the integral test and the comparison test, from which the root test and ratio test are derived. Finally, a more refined "Raabe’s test" is usually introduced. The precision (strength) of these methods varies, and it is generally believed that Raabe’s test has the widest range of applications. However, in my view, the comparison test based on the p-series is already sufficient for all problems; it is truly the most powerful method.

The p-series is the familiar: \sum_{n=1}^{\infty} \frac{1}{n^p}

Through the integral test, it can be shown that the series converges when p > 1 and diverges otherwise. Although I cannot prove it, I believe the following conclusion holds:

If a series with positive terms \sum_{n=1}^{\infty} a_n converges, one can always find a constant A and a constant p > 1 such that for every term, a_n < \frac{A}{n^p}.

Since the convergence of the p-series is proven via the integral test, the most fundamental method should be the integral test itself. However, it is not convenient to apply, so we will only discuss the "comparison test based on the p-series."

Below is an example to illustrate how to find A and p.

\sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!} \frac{1}{2n+1}

Here, !! denotes the double factorial, i.e., n!! = n(n-2)(n-4)\dots. The above series is convergent, but its conditions are quite strong, making general convergence tests inapplicable. Textbooks usually use Raabe’s test to judge it, indicating that it is a series with a relatively slow convergence speed. Below, we use the comparison test to prove its convergence.

Suppose \frac{(2n-1)!!}{(2n)!!} \frac{1}{2n+1} < \frac{A}{n^p}

We primarily consider mathematical induction. That is, for n+1, we have: \frac{(2n+1)!!}{(2n+2)!!} \frac{1}{2n+3} < \frac{A}{(n+1)^p}

Dividing the two inequalities, a sufficient condition for the former to imply the latter is: \left(\frac{2n+1}{2n+2}\right)\left(\frac{2n+1}{2n+3}\right) < \left(\frac{n}{n+1}\right)^p

It is always possible to choose a sufficiently small p that is still greater than 1 to make the above inequality hold consistently. Let’s first estimate p. We rewrite it as: \left(\frac{2+\frac{1}{n}}{2+\frac{2}{n}}\right)\left(\frac{2+\frac{1}{n}}{2+\frac{3}{n}}\right) < \left(\frac{1}{1+\frac{1}{n}}\right)^p

Expanding both sides in terms of \frac{1}{n} (note that we are currently estimating, not yet proving): 1-\frac{1.5}{n} < 1-\frac{p}{n}

A preliminary estimate is that p=1.5 suffices. Of course, to be safe, one could also take p=1.4, 1.3, etc. But the fact is that when p=1.5, the inequality \left(\frac{2n+1}{2n+2}\right)\left(\frac{2n+1}{2n+3}\right) < \left(\frac{n}{n+1}\right)^p already holds consistently (the proof is omitted here, but it is quite straightforward). How should we choose A? In fact, once p is chosen, we only need to select a sufficiently large A such that the first term of \frac{A}{n^p} is greater than the first term of the original series. In this case, it is simple; taking A=1 is sufficient. Thus, our proof is complete. By constructing \frac{(2n+1)!!}{(2n+2)!!} \frac{1}{2n+3} < \frac{1}{(n+1)^{1.5}} the proof is finished.

On the other hand, proving divergence is even easier because we do not need to choose a suitable p (we can use p=1); we only need to pick an appropriate A!