Summation of series containing logarithm: $\sum_{n=1}^\infty \ln \frac{(n+1)(3n+1)}{n(3n+4)}$

Question

How do I find the sum of the series: $$\ln \frac{1}{4} + \sum_{n=1}^\infty \ln \frac{(n+1)(3n+1)}{n(3n+4)} $$

I tried expanding the terms on numerator and denominator and got $$\ln \frac{1}{4} + \sum_{n=1}^\infty \ln\left(1+\frac{1}{n(3n+4)}\right)$$ but I am stuck as I couldn't figure out how to proceed further.

Answer 1

This is a telescopic series. You may write $$ \ln \frac{(n+1)(3n+1)}{n(3n+4)} =\left[\ln(\color{blue}{n+1})-\ln (\color{blue}{n})\right] -\left[\ln (3(\color{purple}{n+1})+1)-\ln (3\color{purple}{n}+1)\right] $$ giving $$ \sum_{n=1}^N\ln \frac{(n+1)(3n+1)}{n(3n+4)} =\left[\ln(\color{blue}{N+1})-\ln \color{blue}{1}\right]-\left[\ln (3(\color{purple}{N+1})+1)-\ln\color{purple}{ 4}\right] $$ or

$$ \sum_{n=1}^N\ln \frac{(n+1)(3n+1)}{n(3n+4)} =-\ln \left(3+\frac1{N+1}\right)+2\ln 2 $$

and you may conclude easily.