<s>[INST] How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$ $$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the squeeze theorem by putting the $a_n$ between $\frac{1}{(2n-1)(2n+1)}$ and $\frac{1}{(4n-1)(4n+1)}$ but it doesn't work.
 [/INST] Alternatively one may recall the series representation for the digamma function
$$\begin{equation} 
\psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}  
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant. 
Then by partial fraction decomposition we have
$$
\begin{align}
\frac1{(6n-3)(6n+2)} &= \frac15\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)\\\\
&=\frac1{30}\left[\left(\frac{1}{n-1/2}-\frac1n\right)-\left(\frac{1}{n+1/3}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac1{(6n-3)(6n+2)} =\frac1{30}\left(\psi\left(\frac43\right)-\psi\left(\frac12\right)\right)
$$ equivalently

$$
\sum_{n=1}^{\infty}\frac1{(6n-3)(6n+2)} =\frac{1}{180} \left(18-\sqrt{3} \pi -9 \ln 3+24 \ln 2\right)
$$ 

where we have used special values of the digamma function.
</s>