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Without getting bogged down in details, I'll list the names only (btw, the tests in each group aren't in any particular order) It seems that the strategy I generally use is this:

  1. Divergence test first

  2. Is it a recognizable form? p-series or geometric?

  3. a) No negative terms? Integral, direct comparison, limit comparison?
    b) Possibly negative terms? Alternating series test, root test, ratio test?

And then each test in the category has its own favourite type of series to work with (like ratio test with factorials and nth powers, for example)

This is hardly a flow chart but I do like the logic of making sure to first check if the series would diverge from the very beginning, and then check if the series is anything familiar. After that, the tests differ in criteria concerning positive and negative signs of the terms.

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  • $\begingroup$ You asked practically the same on Mathematics Educators not quite two days ago. $\endgroup$ – Daniel Fischer Aug 8 '20 at 11:08
  • $\begingroup$ Yes, I did, but I think feedback from this community would be valuable too since the Math stackexchange usually involves homework help involving series. Looking at various answers to questions, the answers often given the most helpful test to use in the first lines. I want to see the reasoning behind it and try to figure out what is it that answer-givers see that question-askers don't. Are they thinking through the tests or are they answering the questions based on analogy. But even then, analogy suggests that they have solved did similarly related series, and so how did they solve them? $\endgroup$ – Robbie_P Aug 8 '20 at 13:05
  • $\begingroup$ The root and ratio tests are for nonnegative series. $\endgroup$ – zhw. Aug 8 '20 at 14:47
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Kill the thing, then think, how to cook it

So, only heavy machinery like limit comparison test and integral test. For limit comparison test you need to memorize things like $$\sum\limits_n \frac{1}{a^n},\, \sum\limits_n \frac{1}{n^p},\, \sum\limits_n \frac{1}{(\ln n)^p},\,\hbox{etc.} $$ So, first: see asymptotical behaviour -- throw away any sines (if it's not Flint-Hill series), cosines, lower terms and see with what to compare. Likely $90-95\%$ of summations will go.
If (after simplifying) can integrate, if rather simple -- integrate. If no, then think. And also any familiar Taylor series will do if pops up.

Why

If you're at home with internet and wolframalpha available, it says what test to use, so the OP question will not be raised anyway. So it's an exam or contest -- thus time limited -- thus no time for tests, giving $1$ (i.e. "inconclusive"). But, when you killed a thing, you'll have to present ("cook") it to others -- to write the solution. Then you may want to choose a "lighter" test if you recall some -- like ratio test or root test or smth.

So I will demonstrate how it works -- there are "related" column on the right. $$\hbox{1. }\sum\limits_{n=1}^\infty \sin^2\left(\frac{\pi}{n}\right) :\sin(x)\sim x\hbox{ when }x\to 0\hbox{, so }\sim\frac{1}{n^2}\hbox{ -- converges}$$ $$\hbox{2. }\sum\limits_{n=1}^\infty \frac{1+5^n}{1+6^n}:\ \frac{1+5^n}{1+6^n}\sim\left(\frac{5}{6}\right)^n\hbox{ -- converges}$$ $$\hbox{3. }\sum\limits_{n=1}^\infty nx^{n+1}\hbox{ -- derivative of Taylor series of }\frac{1}{1-x}\hbox{, so ratio test}\\ \hbox{ and very accurate on the borders after}$$ $$\hbox{4. }\sum\limits_{n=1}^\infty (-1)^n \ln(n):\ \lim\limits_{n\to\infty}(-1)^n \ln(n)\ne 0\hbox{ -- diverges}$$ $$\hbox{5. }\sum\limits_{n=1}^\infty \frac{\sin\left(n-\sqrt{n^2+n}\right)}{n}\hbox{ -- how did it get the score of 28? ... :}\\ \sin\left(n-\sqrt{n^2+n}\right)= \sin\left(\frac{\left(n-\sqrt{n^2+n}\right)\left(n+\sqrt{n^2+n}\right)}{n+\sqrt{n^2+n}}\right)\sim\sin\left(\frac{1}{2n}\right)\sim \frac{1}{2n}\hbox{ -- so overall converges}$$

P.S.

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