I know only two convergence tests for non positive series Dirchlet's and Abel's. Both can only confirm the convergence.

Is there a test that can show if a non positive series diverges ? I expected it is complicated since it isn't thought in school.

By non positive I mean a series that doesn't have only positive terms. Example $\sum\frac{\sin n} n $

Due to the votes for closing the question I want to say that I am asking for an algorithm that shows if a series converges or not, for positive series Gauss's test is an example. The question is about any series in general, so I cannot make the question more specific as suggested by the votes.

  • 1
    $\begingroup$ There is the test that its term doesn't tend to zero. $\endgroup$
    – user763871
    Apr 3, 2020 at 19:57
  • $\begingroup$ ~~Sometimes you can construct an different series by reordering groups of summations such that you end up with a positive term series.~~ Only valid if series is bsolutely convergent, good point. $\endgroup$
    – TrostAft
    Apr 3, 2020 at 19:59
  • $\begingroup$ @TrostAft i cannot reorder the terms if series isn't absolute convergent $\endgroup$
    – Milan
    Apr 3, 2020 at 20:01
  • $\begingroup$ When you say "non-positive" do you mean a series whose terms are negative (or zero) or one where there is a mixture of positive or negative? $\endgroup$
    – Henry
    Apr 3, 2020 at 20:01
  • 2
    $\begingroup$ I might say it is a mixture of positive and negative terms. You might look at the sequence of partial sums. $\endgroup$
    – Henry
    Apr 3, 2020 at 20:03

1 Answer 1


Convergence of positive series (especially with decreasing terms) is largely about asymptotics: do the terms decay fast enough to make the sum finite? This is often straightforward because the sum of positive terms of some size is roughly their number times their size. On the other hand, convergence of conditionally convergent series is about cancellation: do positive and negative terms together add up so something small? This is much more subtle, because you are trying to show that a sum of terms of some size is much smaller than the number of terms times the typical size. There is no general rule and you often have to be clever.

The main tool for understanding oscillation is integration by parts, or its discrete cousin summation by parts (also known as Abel summation). In a series $\sum_n a_n b_n$ you replace the first factor with its discrete derivative $a_n-a_{n+1}$ and the second factor with its discrete integral, the partial sum $B_n = b_1 +\dots b_n$, giving the equivalent series $\sum_n (a_n-a_{n+1}) B_n$. The general ideal is that if $b_n$ is oscillatory then $B_n$ will tend to be much smaller because of cancellation (e.g. $\sum_{k=1}^n \sin k$ are uniformly bounded in $n$ despite the typical summand having magnitude about $1$) while the differences $a_n - a_{n+1}$ might also be small (because the $a_n$ are small and decaying).

So how do you detect non-cancellation? If it happens on a term-by-term basis, you can just group the elements. Consider the series with terms $a_n = \begin{cases} 1/n & n\textrm{ odd} \\ -1/2n & n\textrm{ even}\end{cases} $. For $n$ odd the sum of the term and the consecutive one is $\frac1n-\frac1{2(n+1)}\sim \frac{1}{2n}$ so the even partial sums diverge.

More generally it's hard to give general recipes. Do you have a specific series in mind?


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