$$\sum_{n=1}^{\infty}\frac{1}{n^2-ln(n)}$$ I was comparing it to $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$ and I had limit $$\lim_{n\to \infty}\frac{a_n}{b_n}$$ where $${a_n}=\frac{1}{n^2-ln(n)}$$ and $${b_n}=\frac{1}{n^2}$$ but at the end I have that series diverges, but it should converge instead.
Thanks for helping in advance!
For each positive integer $ n \geq 2 $, we have $ \ln{n} < n $, and thus $$ \frac{1}{ n^2 - \ln{n} } < \frac{1}{ n^2 - n } = \frac{1}{n-1} - \frac{1}{n}. $$ And we have $$ \sum_{ k = 2 }^n \left( \frac{1}{k-1} - \frac{1}{k} \right) = 1 - \frac{1}{n} ,$$ so $$ \sum_{n=2}^\infty \frac{1}{ n^2 - n } = 1 .$$ Finally, we have \begin{aligned} &\sum_{n=1}^\infty \frac{1}{ n^2 - \ln{n} } \\ =& 1+\sum_{n=2}^\infty \frac{1}{ n^2 - \ln{n} } \\ \leq& 1 + \sum_{n=2}^\infty \frac{1}{ n^2 - n } \\ =& 2. \end{aligned}
It seems as though you were trying to use the limit comparison test (LCT) with $${a_n}=\frac{1}{n^2-ln(n)}$$ and $${b_n}=\frac{1}{n^2}.$$
It seems that, for some reason you haven't told us, upon application of the LCT you ended up showing that the series diverges. But this is not true, and so you must have made a mistake before you concluded that the series diverges. Since you haven't shown us your working that made you conclude that the series diverges by the LCT, we cannot yet help show you where you went wrong in your working. However, I can show you how the LCT does indeed show that the series converges.
Since $a_n\geq 0, b_n\geq 0\quad \forall\ n\in\mathbb{N},\ $ and
$$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^2}{n^2-ln(n)} = \lim_{n\to\infty} \frac{\left(\frac{1}{n^2}\right) n^2}{\left(\frac{1}{n^2}\right)\left(n^2-ln(n)\right)} = \lim_{n\to\infty} \frac{1}{1-\frac{ln(n)}{n^2}} = 1,$$
correct usage of the LCT tells us that $\displaystyle\sum_{n=1}^{\infty} a_n$ converges $\ \iff \displaystyle\sum_{n=1}^{\infty} b_n$ converges. I'm assuming you already know and can use the result that $ \displaystyle\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. We can therefore conclude by LCT that $ \displaystyle\sum_{n=1}^{\infty} a_n$ converges.