In search of a "perfect" test on (positive) series convergence

Question

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!

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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)

Answer 1

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.

Answer 2

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?

Answer 3

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.

Answer 4

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 $h\le1$.

Answer 5

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 $(0,1]$.

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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 $$ \int_0^1 t^{-1/\alpha}dt < \infty \iff \frac{1}{\alpha} < 1 \iff \alpha > 1. $$

Answer 6

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.

Answer 7

If a perfect test existed, we could decide the convergence or divergence of every sequence.

Indeed, every series is a sequence, namely the sequence of partial sums of its general term. Conversely, if $\{a_n\}_{n\ge 0}$ is any sequence, then we can see $\{a_n\}_{n\ge 0}$ as a series by defining $u_0=a_0$ and $u_n=a_n-a_{n-1}$ for $n\ge 0$ so that

$$ a_n = \sum_{k=0}^n u_k $$

is the partial sum of the series of general term $u_n$. If we have a perfect test to decide the convergence of $\sum_{n\ge 0}u_n$, we would have (in theory) a perfect test for the convergence of every sequence.

Intuitively, there is no method to determine the convergence of every sequence, so a perfect test is extremely unlikely to exist.