In order to evaluate this limit :
$$\lim_{n\to \infty} \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$$
I have to compute the following sum :
$$ \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$$
$$1-\frac{2}{k(k+1)}=\frac{k^2+k-2}{k^2+k}$$
So :
$$\log_{\frac{1}{3}} \left(\frac{k^2+k-2}{k^2+k}\right)=\log_{\frac{1}{3}}\left(k^2+k-2\right) -\log_{\frac{1}{3}} \left(k^2+k\right)$$
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 $$\lim_{n\to \infty} S_n =1$$
Should I really use "Gamma function" to solve this 12th grade sum ? Or is there any other method ?
Best Answer
In this answer I'm considering $b=\frac{1}{3}$.
\begin{align} \sum_{k=2}^n \log_b\left(1-\frac{2}{k(k+1)}\right)&= \sum_{k=2}^n \log_b\left(\frac{k^2+k-2}{k(k+1)}\right)\\ &=\sum_{k=2}^n \log_b \left(\frac{k+2}{k+1}\right) +\log_b \left(\frac{k-1}{k}\right)\\ &=\sum_{k=2}^n \log_b (k+2)-\log_b(k+1) +\sum_{k=2}^n \log(k-1)-\log_b(k)\\ &=\sum_{k=2}^n \log_b (k+2)-\log_b(k+1) -\sum_{k=2}^n \log(k)-\log_b(k-1)\\ &=\log_b(n+2)-\log_b(3) -\log_b(n)+\log_b(1)\\ &=\log_b\left(\frac{n+2}{n}\right)-\frac{\ln(3)}{\ln(1/3)}\\ &=\log_b\left(\frac{n+2}{n}\right)+\frac{\ln(3)}{\ln(3)}\\ &=\log_{\frac{1}{3}} \left(1+\frac{2}{n}\right)+1\\ \lim_{n\to \infty} \sum_{k=2}^n \log_b\left(1-\frac{2}{k(k+1)}\right)&=\lim_{n\to \infty} \log_{1/3}\left(1+\frac{2}{n}\right)+1\\ &=1 \end{align}
Recall : $$\frac{1}{n} \overset{n\to \infty}{\longrightarrow} 0 \ \ \ \ \ \ \text{;} \ \ \ \ \ \ \log_a(1)=0 \ \ \ \ \ \ \text{;}\ \ \ \ \ \ \log_b (x)=\frac{\ln x}{\ln b}\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ln(1/a)=-\ln(a)$$