sequences and series – How to evaluate the following limit \$lim_{nto infty} sum_{k=2}^nlog_{frac{1}{3}} left(1-frac{2}{k(k+1)}right)\$?

In order to evaluate this limit :
$$lim_{nto infty} sum_{k=2}^nlog_{frac{1}{3}} left(1-frac{2}{k(k+1)}right)$$
I have to compute the following sum :
$$sum_{k=2}^nlog_{frac{1}{3}} left(1-frac{2}{k(k+1)}right)$$
$$1-frac{2}{k(k+1)}=frac{k^2+k-2}{k^2+k}$$
So :
$$log_{frac{1}{3}} left(frac{k^2+k-2}{k^2+k}right)=log_{frac{1}{3}}left(k^2+k-2right) -log_{frac{1}{3}} left(k^2+kright)$$
I’m really lost now, and the worse is Wolfram Alpha is giving to me this crazy value :

$$S_n=frac{ln (3 Gamma(n+1) Gamma(n+2)-ln (Gamma(n) Gamma(n+3))}{ln 3}$$
And when $$lim_{nto infty} S_n =1$$
Should I really use “Gamma function” to solve this 12th grade sum ? Or is there any other method ?