There are JUST 6 methods in SumConvergence. 7th method can be: https://mathematica.stackexchange.com/a/110324/82985 Essentially if you have `SumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n]` it does not work, but you can create a new series by grouping, in this case grouping in pairs. Like this: SumConvergence[1/(Sqrt[2 n] + 1) - 1/(Sqrt[2 n + 1] - 1), n] And that will allow Mathematica to immediately find this new series diverges. Then simple but a little obscure rule in calculus says that then the original series can be only divergent. See https://en.wikipedia.org/wiki/Series_(mathematics)#Grouping (obscure cause everyone confuses it with Risch theorem, but grouping is not a rearrangement). If you have convergent series its groupings can be only convergent, and if any series is divergent its groupings can be divergent and convergent. That means when you code it you will have learn nothing if you get True for new series and only False means False for original series. And you may want to try 1+2, 3+4, 5+6... element and also 1+2+3, 4+5+6, 7+8+9... element...

BTW, I made a typo when I wrote the post: Then simple but a little obscure rule in calculus says that then the original series can be only divergent. I edited to divergent. If you have convergent series its groupings can be only convergent, and if any series is divergent its groupings can be divergent and convergent.

Wait, no. If grouping is convergent you can say nothing about anything. You need to return the input. The feature I propose does not WORK AT all for convergent series. It is useless for them... As an example this convergent series: ((-1)^n)/(n + (-1)^n) for n from 2 to Infinity. Also, Method -> groupTermsCheck[2] must be HIGHEST priority together with DivergenceTest!!! There are no any requirements on this test just like with DivergenceTest (which is just limit of terms must be 0 for series to be even considered convergent).

Also what happens with 1 - Cos[Pi/n]? It crashes or something?

New Method for SumConvergence: if grouping of series is divergent so is original series