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

Solution 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.

Solution 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?

Solution 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.

Solution 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 $h\le1$.

Solution 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 $(0,1]$.

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