What are the values ​​for which the series converges? Determining the values ​​of $a$ and $b$ so that the series $\sum a_n$ converges, where 
 $$a_n=\ln n-a\ln(n+1)+b\ln(n+2)$$
 A: \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}
$$
1 - a + b = 0\,,
\quad
- a + 2b = 0
\qquad\Longrightarrow\qquad
\color{#ff0000}{\large a = 2\,,\quad b = 1}
$$
A: 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
