What are the values ​​for which the series converges?

Question

Determining the values ​​of $a$ and $b$ so that the series $\sum a_n$ converges, where $$a_n=\ln n-a\ln(n+1)+b\ln(n+2)$$

Answer 1

\begin{align} a_{n} &= \ln\left(n\right) - a\ln\left(n + 1\right) + b\ln\left(n + 2\right) \\[3mm]&= \ln\left(n\right) - a\left[ \ln\left(n\right) + {1 \over n} - {1 \over 2n^{2}} + \cdots \right] + b \left[ \ln\left(n\right) + {2 \over n} - {2 \over n^{2}} + \cdots \right] \\[3mm]&= \left(1 - a + b\right)\ln\left(n\right) + \left(-a + 2b\right){1 \over n} + \left({1 \over 2}a - 2b\right){1 \over n^{2}} + \cdots\,, \qquad n \gg 1 \\[5mm]& \end{align}

$$ 1 - a + b = 0\,, \quad - a + 2b = 0 \qquad\Longrightarrow\qquad \color{#ff0000}{\large a = 2\,,\quad b = 1} $$

Answer 2

Assuming that $b$ is an integer, $$\begin{multline} a_n=\ln n-a\ln(n+1)+b\ln(n+2) = \ln\left(\frac{n(n+2)^b}{(n+1)^{a}}\right)= \\ = \ln\left(\frac{n((n+1)+1)^b}{(n+1)^{a}}\right)=\ln\left(\frac{n\sum_{k=0}^{b}{b\choose k}(n+1)^k}{(n+1)^{a}}\right). \end{multline}$$ If $a=b+1$ then $$ a_n \to \ln(1) = 0. $$ Now there is still the issue of "which" integers $b$ are acceptable.

Thanks to Andres Caicedo for pointing out some inconsistencies