Tips on intuition for solving these convergent/divergent series problems? I have really bad intuition when figuring out what test/approach I should use for checking whether a series converges. This definitely stems for my weak foundation in math overall but specifically algebra/rearranging equations; I can't visualize quickly that one equation "looks like" another. Could you give me some tips on how you would approach these?
Example 1:
I initially started this as a ratio test but when I simplified it I ended up getting divergent as the answer which is wrong...
$$
A_n=\frac{n}{n+\sqrt{n}}
$$
Example 2:
The question itself said to use the integral test, but I probably wouldn't have thought of it myself. Furthermore my answer ended up as $\infty - \infty$ and I wasn't sure what this would mean $DNE\implies divergent$ (doubt that's right in the first place).
$$
A_n=\sum^{\infty}_{x=1} \frac{ln(x)}{x}
$$
Example 3: I thought this was the use of the addition rule where convergent + conv/divergent $\implies convergent$ but turns out the answer was divergent. Does the addition rule not apply for subtraction??
$$
A_n=\sum^{\infty}_{x=1}\frac{1}{x}-\frac{1}{x^2}
$$
If I'm allowed, I will edit this question with a few more examples so that I can really drill things home but idk if that's not allowed so holding back for now.
ALSO, to find what it converges to we've only used the limit as $n \to \infty$ of the $S_n=\frac{A_0(1-r^k)}{1-r}$formula, where we get $r$ by calculating a few sample values and dividing $A_{n+1}/A_n$ but that doesn't seem to be working as I get to questions with very obscure looking $A_n$. Is there another way (at the undergraduate Calc.3 level)?
 A: *

*Do the terms converge to $0$? E.g. $\frac {n}{n+\sqrt n}=\frac {1}{1+1/\sqrt n}\to 1.$ 

*Are the terms  larger (eventually) than the terms of a positive series that you know to be divergent? E.g. $\frac {\ln n}{n}>\frac {1}{n}$ when $n\ge 3.$ And if $n\ge 4$ then $\frac {n}{n+\sqrt n}\ge \frac {n}{n+(n/2)}=2/3.$

*Are the terms comparable to $\frac {p(n)}{q(n)}$ for some polynomials $p(n),q(n)$? If $p\ne 0$ then $\sum_n\frac {p(n)}{q(n)}$ converges iff $deg(q)\ge 2+deg(p).$

*If $\sum_n B_n$ converges absolutely then $\sum_n(A_n+B_n)$ converges iff $\sum_n A_n$ converges. E.g. $A_n=\frac {1}{n}$ and $B_n=-\frac {1}{n^2}.$ For this example we can also use 2., as $\frac {1}{n}-\frac {1}{n^2}=\frac {p(n)}{q(n)}$ with $p(n)=n-1$ and $q(n)=n^2.$

*The Cauchy Condensation Test. If $a_n\ge a_{n+1}\ge 0$ for all but finitely many $n $ then $\sum_na_n$ converges iff $\sum_n2^na_{2^n}$ converges. E.g.$\sum_{n\ge 2}\frac {1}{n(\ln n)^k}$ converges iff $\sum_{n\ge 2}\frac {1}{n^k(\ln 2)^k}$ converges... and  by a 2nd application of this test, $\sum_{n\ge 2}\frac {1}{n^k}$ converges iff $\sum_{n\ge 2}\frac {2^n}{2^{nk}}=\sum_{n\ge 2}\frac {1}{2^{n(k-1)}}$ converges iff $k>1....$ so $\sum_{n\ge 2}\frac {1}{n(\ln n)^k}$ converges iff $k>1.$


*Alternating series. If $a_n\ge a_{n+1}\ge 0$ for all but finitely many $n,$ and if $a_n\to 0$ then $\sum_n(-1)^n a_n$ converges. E.g. $a_n=1/n.$
