# How to show a non positive series diverges? [closed]

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.

• There is the test that its term doesn't tend to zero.
– user763871
Apr 3, 2020 at 19:57
• ~~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. Apr 3, 2020 at 19:59
• @TrostAft i cannot reorder the terms if series isn't absolute convergent Apr 3, 2020 at 20:01
• 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? Apr 3, 2020 at 20:01
• I might say it is a mixture of positive and negative terms. You might look at the sequence of partial sums. Apr 3, 2020 at 20:03

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.