# Can we always establish whether an infinite series converges or diverges?

I'm currently working with infinite series for my calculus class, and I'm wondering whether we always (in theory) can establish whether a series is divergent or convergent? Of course, it might be computationally hard, but are there a class of series where we simply lack the tools to determine whether the series converges or diverges?

• – Martin R Feb 25 '18 at 9:51
• Or Series whose convergence is not known on MathOverflow. – Martin R Feb 25 '18 at 9:52
• I don't have an example, but it's probably possible to define a series that you'd have to be able to solve the halting problem to determine its convergence. – Shufflepants Feb 25 '18 at 22:09

I'm not sure whether this is what you want or not. Consider the series $$\displaystyle\sum_{n=1}^\infty a_n$$, with$$a_n=\begin{cases}1&\text{ if }2^n-1\text{ is prime}\\0&\text{ otherwise.}\end{cases}$$It is not known whether it converges or not.

• Is this a well-known series used in prime counting and such or did you just make it up on the spot? Thanks. – PatrickT Feb 26 '18 at 6:45
• @PatrickT I made it on the spot. The sum of this series is the number of Meresenne primes. Since we don't know whether there are infinitely many of them or not… – José Carlos Santos Feb 26 '18 at 7:21
• shouldn't it be "It is not known whether"? – snoram Feb 26 '18 at 14:35
• Thanks José Carlos! – PatrickT Feb 26 '18 at 16:04
• @snoram I've edited my answer. Thank you. – José Carlos Santos Feb 26 '18 at 16:27

Just for fun, whatever foundational system $S$ you are working in, as long as $S$ can handle basic arithmetic, here is a series that $S$ (and hence you) cannot prove that it converges, and you hope $S$ never proves that it diverges! $\def\nn{\mathbb{N}}$

Let $f : \nn \to \nn$ such that for each natural $n$, we have $f(n) = 1$ if there is a proof over $S$ of length at most $n$ of a contradiction, and $f(n) = 0$ otherwise.

Then $S$ cannot prove that $\sum_{n=0}^\infty f(n)$ converges (otherwise $S$ proves itself consistent, which contradicts Godel's incompleteness theorem).

And $S$ had better not prove that $\sum_{n=0}^\infty f(n)$ diverges (otherwise $S$ proves itself inconsistent).

A rather curious fact (if one has never seen the incompleteness phenomenon before) is that, if $S$ is consistent, then $S$ can actually prove that $f(0) = 0$ and $f(1) = 0$ and $f(1+1) = 0$ and so on, but cannot prove "$\forall n \in \nn\ ( f(n) = 0 )$".

One can improve this, via something equivalent to Rosser's trick, to obtain a series that $S$ (and hence you) cannot prove or disprove whether it converges!

In general, any question of the form "Is it always possible to determine (prove or disprove) whether an object in collection $C$ satisfies property $P$?" is likely to have "no" as the answer if you can use $P$ on suitable members of $C$ to determine whether any given sentence over $S$ is provable or not.

No, there are some criteria (for example $|a_n|/|a_{n+1}| \rightarrow l<1$) you can sometimes use, but even when you have one of those, on the limit cases (for example if $|a_n|/|a_{n+1}| \rightarrow 1$, for this last criterium), you'll have to prove it "by hand" (meaning there is no general way to do so)...

Every series is convergent or divergent.

However there is no general method to find if any series is indeed convergent or divergent.

There are two reasons.

Some series (most of them) are not expressible in any shortened form to even define them except stating each term, meaning the sum of such series is not known except stating each partial sum, but then you can say nothing about its convergence or divergence.

Extracting the information about convergence means we should be able to extract at least that much about any series. Unfortunately there are series (most of them actually) that you cannot extract any of the global properties. None. Only a couple of first terms and consequently partial sums. Nothing else. Nada, zero, rien, nichts, τίποτα.

For the series whose partial sums are compressible, there is no universal algorithm, so they are recursively undecidable.

The deep reason for this is that divergence and convergence are easily made equivalent to an algorithm halts or does not halt. You can effectively turn any algorithm into a series whose convergence/divergence you want to analyze.

We cannot know which series we can solve by which method. But each particular series that is compressible might (should) have a particular method that would decide if it is convergent or divergent. So each compressible series is likely a world on its own.

So it is triple naked :(