# In search of a “perfect” test on (positive) series convergence

Thus far mathematicians have developed many powerful tests on the convergence of a positive series (I mean $\displaystyle\sum_{i=1}^{\infty}a_i$ specifically), such as :

Cauchy's Test
which deals with the upper limit of $\lambda_n=\sqrt[n]{a_n}$, but comes to nothing when the upper limit is $1$.

D'Alembert's Test
which deals with the upper or lower limit of $\lambda_n=\frac{a_{n+1}}{a_n}$, but comes to nothing when the upper limit $\ge1$ or the lower limit $\le1$.

Raabe's Test
which deals with $\lambda_n= n\Big(\frac{x_n}{x_{n+1}}-1\Big)$, but comes to nothing when $\lim{\lambda_n}=1$.

Bertrand's Test
which deals with $\lambda_n= (\ln n)\Big[n\Big(\frac{x_n}{x_{n+1}}-1\Big)-1\Big]$, but still comes to nothing when $\lim\lambda_n=1$.

$$\vdots$$
And in my books it says, this progress never ends, "...We can always go on and establish a even more powerful test, with more complicated proof, though..." I don't really know how, but since this is not the point of my question, you may just skip it.
Well, anyhow I hope you could take a close look at these tests. They are really brilliant in that they can tell you whether the series converges or otherwise based directly on the information of $a_n$, which won't be too obscure. But they are still NOT perfect. They are all "if" type but not "iff" type. I mean, they are all of such pattern:

The series converges if $\lim \lambda_n$ blah blah blah, and diverges if $\lim \lambda_n$ blah blah blah. The worst is, if they both fail us, WE KNOW NOTHING!

How I hope that I could replace "if"s with "iff"s, and get rid of the "we know nothing" case!
So I'm quite wondering whether there is a perfect test that:
(1) is based directly on $\lambda_n$ which $a_n$ gives rise to
(2) is of such pattern as:

The series converges iff $\lim \lambda_n$ blah blah blah, and diverges iff $\lim \lambda_n$ blah blah blah. (Say, we know everything!)

I know there might be only dim hope, but I'm still curious. Any help will be specially appreciated. Best regards!

Further Note
In the first place, I'm sincerely grateful to all the help you guys provide for me. However, I'm afraid I have to make a note here because many answers posted here are not what I'm looking for. Well I'm far from criticizing, but I think I need to perhaps make my question clearer so that I'm not misleading your answers.

The problem is that some answers here do not really meet the requirement (1) mentioned in my question. Please read (1) closely, I want the test to be based directly on $\lambda_n$ to which $a_n$ gives rise, just like the $\lambda_n$s in the tests listed above. In other words, $\lambda_n$ should be immediately accessible via $a_n$, or, $a_n$ gives all the immediate information needed to write out $\lambda_n$. Therefore, $\lambda_n$ is an expression that contains $a_n, a_{n+1}$ etc etc. I don't want to involve the partial sum in my test, nor am I looking for something like a powerful comparison test, because the knowledge of $a_n$ usually cannot enable us to gain the knowledge of the partial sum, or to find another $b_n$ to compare to. In short, I desire something that is based only and immediately on $a_n$. Thanks.

(And, apologies if my post should look too wordy. I'm not a skillful English user)

• I've also wondered this, but maybe it's the case that there's no "nice" answer that also catches every convergent series – dalastboss Mar 4 '15 at 3:13
• The integral test is an "iff" test. – Paul Mar 4 '15 at 3:17
• @dalastboss I know it would be unlikely to develop a theorem that catches every convergent series, but there perhaps (in dim hope) might be one that catches every positive series, I suspect. – Vim Mar 4 '15 at 3:19
• @Paul Yes it is "iff" but it is not perfect. Because It doesn't deal directly with $a_n$ and is not very doable. (Only when $a_n>0$ and $a_n$ is monotonically decreasing can we use the information of $\int_{N}^{\infty} f(x) dx$ to probe the convergence of the series. Even so it is not based directly on $a_n$). – Vim Mar 4 '15 at 3:23
• Vim. Did you check out the source:(dcs.fmph.uniba.sk/bakalarky/obhajene/getfile.php/…)? It is a solid review. – Mark Viola Mar 4 '15 at 5:04

Kummer's test: Let $a_k\ge0$.

1. $\sum a_k$ converges if and only if there exist a positive sequence $p_k$ and $c>0$ such that $$p_k\,\frac{a_k}{a_{k-1}}-p_{k+1}\ge c\quad\forall k\text{ large enough.}$$
2. $\sum a_k$ diverges if and only if there exists a positive sequence $p_k$ such that $\sum1/p_k$ diverges and $$p_k\,\frac{a_k}{a_{k-1}}-p_{k+1}\le0\quad\forall k\text{ large enough.}$$

Most of the usual tests are derived by using an appropriate choice of the sequence $p_k$.

Reference

J. Tong, Kummer's Test Gives Characterizations for Convergence or Divergence of all Series, American Mathematical Monthly 101 (1994) 450-452.

• In the convergence case, is $\{p_n\}$ free to choose? – Vim Mar 4 '15 at 11:21
• Ok. I got it. So all we need is a fitting $p_n$. Thanks. – Vim Mar 4 '15 at 11:27
• Shouldn't it be $\frac{a_k}{a_{k+1}}$ instead of $\frac{a_k}{a_{k-1}}$? – Sil Aug 8 '18 at 18:42

There is no "perfect" test. One of the reasons is that, no matter how slowly the sum of a series converges, there is another series whose sum converges more slowly.

There is a good discussion of that here:

Can a sequence which decays more slowly still yield a converging series?

• Thanks. But I'm sorry I don't quite understand the relation between your "reason" and the absence of a perfect test. Could you gimme a hint on this? – Vim Mar 4 '15 at 3:30
• My guess is that you mean whatever powerful test we find, there will always be some tricky examples that even this test can do nothing with, is it? – Vim Mar 4 '15 at 9:13
• @Vim: If memory serves, Baby Rudin (in the exercises) has some comments along the lines of marty cohen's answer, to the effect that there exists no "universal comparison series". To get a definitive answer to your question (e.g., a non-existence theorem), I expect you need to formalize the definition of a "convergence test". Lots of people on site are sufficiently expert in "foundational" issues to help with that sort of task. (I am not one of them.) – Andrew D. Hwang Mar 4 '15 at 13:51
• @user86418 I found Rudin here. – Vim Mar 4 '15 at 14:29

If $\{a_n\}$ is a sequence of positive numbers, then $$\sum_{n=1}^\infty a_n$$ exists if and only if the sequence of partial sums $$s_N := \sum_{n=1}^N a_n$$ is bounded above. That is the best "iff" that you're going to get. Otherwise, we would have a different definition for convergence of a series.

• It is "iff". But it doesn't meet (1). The "perfect" test I expect is not only "iff" but also "doable". Say, it should be based directly on the information of $a_n$, but not the whole series itself. – Vim Mar 4 '15 at 3:26
• I have edited my question. You can take another look at it if you don't mind. – Vim Mar 4 '15 at 9:08
• I wasn't trying to be flippant. Just pointing out that if there was a "perfect" test then we would be able to use that as the definition for convergence. – Math1000 Mar 4 '15 at 11:31
• Thanks! I know what you mean. What I thought is that perhaps such a test does exist but still doesn't appear yet. Now I begin to doubt it slightly, though – Vim Mar 4 '15 at 11:45
• Not disputing your conclusion ("there is no 'better' convergence test"), but making a good-natured note that your last claim ("Otherwise we would have a different definition...") may be debatable. For example, Riemann integrability is defined in terms of upper and lower sums, but is equivalent to "bounded, and discontinuous on a set of measure zero". Despite the practical utility of the latter, however, we have not changed the definition of integrability. Definitions are often (rightly) attuned to "conceptual correctness", even if they're clumsy for everyday use. :) – Andrew D. Hwang Mar 4 '15 at 13:58

One test that is very powerful that I didn't see mentioned is Gauss's test. Here, if the positive series $\sum a_n$ is such that $$\frac{a_n}{a_{n+1}} = 1+\frac{h}{n}+O(\frac{1}{n^\alpha}),$$for $\alpha>1$, then $\sum a_n$ converges if $h>1$ and diverges if $h\le1$.

• Thanks But I think it might well be a variant version of Raabe 's Test – Vim Mar 4 '15 at 4:37
• It is more general than Rabbe's test. Check out this source. dcs.fmph.uniba.sk/bakalarky/obhajene/getfile.php/… – Mark Viola Mar 4 '15 at 4:44
• It is more general than Rabbe's test. There is no $h$, for which the test fails. That is an improvement over the other tests. Check out this source. dcs.fmph.uniba.sk/bakalarky/obhajene/getfile.php/… – Mark Viola Mar 4 '15 at 4:50
• Oops! I'm reading the beginning and it looks incredible! Is it your work? – Vim Mar 4 '15 at 5:14
• Vim. Kummar is a powerful test from a theoretical view point. But the challenge of finding the $p_n$ that meet the criteria is as challenging as finding a convergent series that dominates the series of interest. That is the reason that I referenced Gauss's test. In practice, one can typically find the expansion required therein. – Mark Viola Mar 5 '15 at 0:32

It is well known that $\sum a_n$ converges iff $a_n$ tends quickly enough to $0$. Another way to look at it is to ask that, for all $t > 0$, the number $$\lambda(t) = \left|\{n \geq 1 : a_n \geq t\}\right|$$

of terms of the sequence greater than $t$ is finite and does not tend too quickly to $+\infty$ as $t$ tends to $0$.

Now, there is a quantitative version of this test: $$\boxed{\displaystyle\sum_{n\geq 1} a_n < \infty \iff \int_0^\infty \lambda(t)dt < \infty}$$ Notice that if $\lim a_n = 0$, it's ok to check the integrability only on $(0,1]$.

Example (Riemann): $a_n = 1/n^\alpha$ with $\alpha > 0$. Then $\lambda(t) = |\{n \geq 1 : n \leq t^{-1/\alpha}\}|$, hence $$\lambda(t) = \begin{cases}0 & \text{if } t > 1\\ \left\lfloor t^{-1/\alpha}\right\rfloor & \text{otherwise} \end{cases}$$ and we know (we can actually compute a primitive) that $$\int_0^1 t^{-1/\alpha}dt < \infty \iff \frac{1}{\alpha} < 1 \iff \alpha > 1.$$

• Thanks! Although I think in the general cases when $t$ tends to zero $\lambda(t)$ will be too complicated to compute, it's still a brilliant "iff" result! – Vim Mar 4 '15 at 14:24
• @Vim: Of course, but an exact computation of $\lambda(t)$ is usually useless: to prove convergence you need only an upper bound, to prove divergence you need only a lower bound. – Siméon Mar 4 '15 at 14:27
• Oh yes, I didn't realize this.. BTW, what's the official name of this powerful test? I'd like to search for more detail on google. – Vim Mar 4 '15 at 14:37
• @Vim: I'm not aware of any name for this test, actually I made it up just for you. Based on how to prove it, you could call it Fubini's test. Or the counting test, maybe? – Siméon Mar 4 '15 at 14:55
• Then I must thank you especially for this :) – Vim Mar 4 '15 at 14:56

A perfect test ($C$) would be an equivalence:

$S=\sum a_n$ converges if and only if $a_n$ satisfies a given property $C$.

One such test is the Cauchy test: $S_n$ converges iff $(\forall \epsilon>0)(\exists n \in \mathbb{N}) (\forall m,k \ge n) (|S_m-S_k| \le \epsilon)$.

Also if $a_n$ is given then we automatically gain knowledge of the partial sums $S_n$.

Another one (for positive series):

$\sum a_n$ converges iff there exists another sequence $(\lambda_n)_{n \ge 1}$ such that $\lambda_n>a_n$ and $\sum \lambda_n$ converges.

• This is not the Cauchy test in the usual sense but the definition. Plus, this is not based on the $\lambda_n$ which $a_n$ gives rise to. (Please note the $\lambda_n$s in these tests I listed) – Vim Mar 4 '15 at 4:27
• @Vim Any "perfect" test of convergence can be used as a definition of convergence too (because of the equivalence). I also added another perfect test. – Marko Karbevski Mar 4 '15 at 4:29
• However, knowledge of $a_n$ doesn't ensure that we know the partial sum as well. In fact in most cases the general term may be simple but the partial sum is just inaccessible to us. For example, given $a_n=\frac1{n^2}$, I don't think anybody would claim that he knows the partial sum "automatically ". – Vim Mar 4 '15 at 6:48
• @Vim I'm afraid such a test does not exist for the simple reason that once you get an equivalence with the definition (or in your words, a perfect test), it gets "almost as hard" (or sometimes even "harder") to prove convergence/divergence using the new test as it is using the basic definition. Also note that sometimes the limit of $\lambda_n$ is very hard to compute if you go by for example Cauchy root test (or any other test that you mentioned). – Marko Karbevski Mar 4 '15 at 10:13
• All these tests that are being used here with $\lim \lambda_n$ are actually comparisons with other sequences. Here is a nice bachelor thesis that i found, that contains quite a few tests: dcs.fmph.uniba.sk/bakalarky/obhajene/getfile.php/… . – Marko Karbevski Mar 4 '15 at 12:09