Integral test to find constant for which series is convergent Problem statement:

Use the integral test to find the value of $b$ for which the following series is
  convergent
$$ \sum_{r=2}^{\infty}\frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)}.$$ 

I evaluated the corresponding integral, which gave the following limit,
$$\lim_{m \to \infty}\left\{ \frac{8}{3b}\ln\left( \frac{mb+1}{2b+1} \right)+\frac{3}{2}\ln\left(\frac{3}{m+1}\right)+\frac{1}{6}\ln(m-1) \right\}.$$
I tried a few different things from there, but with no luck. How do I find the value of $b$ for which this limit exists?

Hint $\sum_{r=2}^{\infty}\frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)}$ is convergent implies 
$$\lim_{r \to \infty} \frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)} =0$$
This yields $b =\mbox{ something}$.
You can then use the integral test to decide if the series converges for this value of $b$.
Alternately
$$\frac{8}{3b}\ln\left( \frac{mb+1}{2b+1} \right)+\frac{3}{2}\ln\left(\frac{3}{m+1}\right)+\frac{1}{6}\ln(m-1) =\\
\ln \left(\frac{\sqrt{27}}{(2b+1)^\frac{8}{3b}} \frac{(mb+1)^\frac{8}{3b}(m-1)^\frac{1}{6}}{(m+1)^\frac{3}{2}}\right)$$
Now compare the powers:


*

*If $\frac{8}{3b}+\frac{1}{6}-\frac{3}{2}>0$
then 
$$\ln \left(\frac{(mb+1)^\frac{8}{3b}(m-1)^\frac{1}{6}}{(m+1)^\frac{3}{2}}\right) \to \infty$$


*

*If $\frac{8}{3b}+\frac{1}{6}-\frac{3}{2}<0$
then 
$$\ln \left(\frac{(mb+1)^\frac{8}{3b}(m-1)^\frac{1}{6}}{(m+1)^\frac{3}{2}}\right) \to -\infty$$
