Hi Everyone! Today we were asked to prove $\displaystyle \sum_{n=1}^{\infty}\frac{\ln(n)}{n^2}$ by comparison test. This is a pretty easy check, since the function is positive, decreasing, and continuous at interval $\displaystyle (1,\infty)$. Integral test tells us that this integral shall converge. But the catch is, what would be the sequence input that shall be compared to the original sequence input so that it can be proven by comparison test. I mean, I can't say that $\displaystyle \frac{\ln(n)}{n^2}$ behaves like $\displaystyle \frac{n}{n^2}$ since $\displaystyle n$ behaves way faster than $\displaystyle \ln(n)$. This comparison also makes the comparison test useless since $\displaystyle \frac{n}{n^2}\geq \frac{ln(n)}{n^2}$. On the other hand, you can't also say that $\displaystyle \frac{\ln(n)}{n^2}$ behaves like$\displaystyle \frac{1}{n^2}$ since $\displaystyle \ln(n)$ is a bit faster than $\displaystyle 1$. So what I was thinking is saying that $\displaystyle \frac{\ln(n)}{n^2}$ behaves like $\displaystyle \frac{verylargenumber}{n^2}$ as $\displaystyle n \to \infty$. The thing is, however, making the numerator a constant will make the series convergent, but in the long run, will never approach $\displaystyle \infty$. So anyone got any idea on what sequence input should I compare the original sequence to, so that I can prove it by Comparison Test. Thanks everyone in advance!