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.
