Compute the sum $ \sum_1^{\infty} \ln{ \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} }$ I need to prove that the sum  $ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) $ converges and equals  $ \ln(\frac{4}{3})$. I tried expanding the fraction into four terms, which cancel each other out, but I'm left with $\ln(4)$. Here's my attempt:
$
\sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) 
= \sum_{1}^{\infty} \ln{(n+1)} - \sum_{1}^{\infty} \ln{n} + \sum_{1}^{\infty} \ln{(3n+1)} -\sum_{1}^{\infty} \ln{(3n+4)} \\
= \sum_{1}^{\infty} \ln{(n+1)} - \sum_{2}^{\infty} \ln{n} + \ln(1) + \sum_2^{\infty} \ln{(3n+1)} + \ln(4) -\sum_{1}^{\infty} \ln{(3n+4)} = \ln(4)$
The last step is simply about playing correctly with the indices and the lower bounds of the series. These telescopic series almost completely cancel each other out, though I can't find the right answer.
Please tell me what I missed.
Thanks

The issue with your calculation is that the sums are not individually convergent.  Therefore, you cannot write
$$\sum_{n=1}^\infty \log \frac{3n+1}{3n+4} = \sum_{n=1}^\infty \log (3n+1) - \sum_{n=1}^\infty \log(3n+4),$$ because each of those sums on the RHS are infinite.
Instead, you need to look at the limiting behavior for a corresponding finite partial sum.  Define
$$f(m) = \sum_{n=1}^m \log \frac{3n+1}{3n+4}.$$  Then $f(m)$ is finite for any positive integer $m$, and now we have no problem computing the telescoping sum:
$$\begin{align}
f(m) &= \sum_{n=1}^m \log (3n+1) - \sum_{n=1}^m \log (3(n+1) + 1) \\
&= \sum_{n=1}^m \log (3n+1) - \sum_{n=2}^{m+1} \log (3n+1) \\
&= \log 4 - \log (3(m+1)+1) \\
&= \log \frac{4}{3m+4}.
\end{align}$$
Similarly,
$$g(m) = \sum_{n=1}^m \log \frac{n+1}{n} = \log (m+1) - \log 1 = \log (m+1).$$
Consequently,
$$f(m) + g(m) = \log \frac{4(m+1)}{3m+4} = \log \frac{4 + 4/m}{3 + 4/m},$$ for all positive integers $m$,
and now taking the limit as $m \to \infty$, we get
$$\sum_{n=1}^\infty \log \frac{(n+1)(3n+1)}{n(3n+4)} = \lim_{m \to \infty} f(m) + g(m) = \lim_{m \to \infty} \log \frac{4 + 4/m}{3 + 4/m} = \log \frac{4}{3}.$$
