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

Let's do it carefully:
$\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\prod_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})$.
Now $\prod_{n=1}^k\frac {n+1}n$ clearly equals $k+1$ as it's just the telescoping series $\require{cancel} \frac {\cancel 2}1\frac {\cancel 3}{\cancel 2}\frac {\cancel4}{\cancel3}.... \frac {\cancel{k-1}}{\cancel{k-2}}\frac {\cancel k}{\cancel{k-1}}\frac {k+1}{\cancel k}$.
[But note $\prod_{n=1}^\infty \frac {n+1}n = \lim_{k\to \infty}(k+1) = \infty$ diverges.  As others have pointed out you can't do an $\lim M\cdot N= \lim N\cdot \lim N = \infty \cdot 0 = ?????$ split which is what you are in essence trying to do.]
$\prod_{n=1}^k \frac {3n+1}{3n+4}= \frac {3n+1}{3(n+1)+1}$ is also a telescopic series. And its product will be $\frac {4}{3(k+1)+1}$ as it is the telescopic seriers $\frac {4}{\cancel 7}\frac{\cancel 7}{\cancel {11}}......\frac {\cancel {3(k-1)+1}}{\cancel{3k+1}}\frac{\cancel{3k+1}}{3(k+1)+1}=\frac 4{3k+4}$.
[Also note $\prod_{n=1}^{\infty} \frac {3n+1}{3n+4} =\lim_{k\to \infty} \frac 4{3k+4} = 0$.  So if we tried to do a $\lim \prod \frac {(n+1)(3n+1)}{n(3n+4)} = \lim (n+1)\frac 4{3k+4} = \lim (n+1)\cdot \lim \frac 4{3k+4}=\infty \cdot 0 = ?????$ it will not work. That is just illegal.]
So $\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})=\ln((k+1)\frac 4{3k+4})=\ln \frac{4(k+1)}{3(k+1)+1}=\ln \frac 4{3+\frac 1{k+1}}$
[But this is acceptable.  If you have an $\lim MN = \lim M \cdot \lim N = \infty \cdot 0$ paradox, it might be fixable if the product $MN$ have terms that cancel out as it does in this case.  $\lim (k+1)\frac 4{3(k+1)+1}\ne \lim(k+1)\lim\frac 4{3(k+1)+1} = \infty \cdot 0=???$ is not okay but $\lim (k+1)\frac 4{3(k+1)+ 1} = \lim \cancel{(k+1)}\frac 4{3\cancel{(k+1)} + \frac 1{\color{purple}{k+1}}}=\frac 43$ is perfectly legal.]
And that's pretty much it:
$\ln(\sum_{n=1}^\infty \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\lim_{k\to \infty}\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\lim_{k\to \infty}\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$ (whoa! careful! is that actually true?  [yes, it is..... so long as the sum converges... which it must else the logs of the partial sums would not converge which we know they do. ])
$\lim_{k\to \infty}\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})=$
$\lim_{k\to \infty}\ln \frac {4}{3+\frac 1{k+1}}=$
$\ln (\lim_{k\to \infty} \frac 4{3+\frac 1{k+1}}=$
$\ln \frac 43$
