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I tried to compare it with $1/n$ which I know diverges:

$$\frac{1}{n\ln^2(n)} \le \frac{1}{n} \Leftrightarrow \\ n\ln^2(n) \ge n \Leftrightarrow \\ \ln^2(n) \ge1\Leftrightarrow \\ \ln(n) \ge 1 \Leftrightarrow \\ n \ge e$$

If $1/n$ diverges and it is greater than the given series for all $n \ge e$... it should diverge, right? Yet the solutions say it converges. Is it because $2 < e$ and $a_2$ is part of the series? Does that make it converge or does that just mean $1/n$ isn't good for comparing?

Note: I could solve this by comparing with $1/\sqrt{n}$ instead but I am still wondering why this failed.

RobPratt
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    I think the integral test is a better bet here. – carmichael561 Feb 23 '20 at 18:26
  • Your comparison is the wrong direction: because $\sum_n 1/n$, diverges you can show divergence of $\sum_n a_n$ if $a_n \ge 1/n$, but you cannot show convergence this way. – RobPratt Feb 23 '20 at 18:28
  • @RobPratt So I can only use the $\lim a_n/b_n$ thing when $a_n \le b_n$, and the $a_n \ge b_n$ thing when $b_n$ diverges? – Segmentation fault Feb 23 '20 at 18:30
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    Your original question and my comment were about the ordinary comparison test. The limit comparison test is different and does not require any inequality between $a_n$ and $b_n$. – RobPratt Feb 23 '20 at 18:32
  • @RobPratt Ok so let me see if I got this right... if $b_n$ diverges and $a_n \ge b_n$, then it also diverges. If $b_n$ converges and $a_n \le b_n$, then a_n converges. If $\lim a_n/b_n = c$, c being a positive finite number, they behave the same way, regardless of what that is, for any $b_n$. Is this correct? – Segmentation fault Feb 23 '20 at 18:35
  • Yes, that is correct. – RobPratt Feb 23 '20 at 20:11

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