How to evaluate the following limit $\lim_{n\to \infty} \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$? 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 ?
 A: The fact that it's meant for 12th-graders is useful information. There can't be too much advanced math involved. It's also useful that you're summing to infinity, not $n$, so let's try to use that. Just take your expression and keep factoring:
$$\log_{1/3}(k^2 +k -2) - \log_{1/3}(k(k+1)) = \\
\log_{1/3}(k+2)+\log_{1/3}(k-1)-\log_{1/3}(k)-\log_{1/3}(k+1)$$
Noting that we began at $k=2$, most of our infinite terms end up canceling out because we have two positive and two negative, and the only nonzero term that is left is
$$-\log_{1/3} 3 = 1.$$
The very answer that you report, and certainly doable for a clever 12th-grader who doesn't know infinitesimal calculus, advanced analysis, etc.
A: 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 :
$$\frac{1}{n} \overset{n\to \infty}{\longrightarrow} 0 \ \ \ \ \ \ \text{;} \ \ \ \ \ \ \log_a(1)=0 \ \ \ \ \ \ \text{;}\ \ \ \ \ \ \log_b (x)=\frac{\ln x}{\ln b}\ \ \ \ \ \ \text{and}\ \ \ \ \ \  \ln(1/a)=-\ln(a)$$
A: $
\newcommand{\l}{\log_{1/3}}
\newcommand{\p}[1]{\left( #1 \right)}
\newcommand{\b}[1]{\color{blue}{#1}}
\newcommand{\r}[1]{\color{red}{#1}}
\newcommand{\g}[1]{\color{green}{#1}}
$
First, notice that
$$1 - \frac{2}{k(k+1)} = \frac{(k+2)(k-1)}{k(k+1)}$$
and the $\log_{1/3}$ of that is
\begin{align*}
\l \p{ \frac{(k+2)(k-1)}{k(k+1)}} 
&= \l(k+2) + \l(k-1) - \l(k) - \l(k+1)\\
&= \l(k-1) - \l(k) - \l(k+1) + \l(k+2) 
\end{align*}
This series telescopes: starting at $k=2$, for instance,
\begin{alignat*}{99}
&\l(1) &&- \b{\l(2)} &&- \r{\l(3)} &&+ \g{\l(4)} \\
+\;&\b{\l(2)} &&- \r{\l(3)} &&- \g{\l(4)} &&+ \l(5) \\
+\;&\r{\l(3)} &&- \g{\l(4)} &&- \l(5) &&+ \b{\l(6)} \\
+\;&\g{\l(4)} &&- \l(5) &&- \b{\l(6)} &&+ \r{\l(7)} \\
+\;&\l(5) &&- \b{\l(6)} &&- \r{\l(7)} &&+ \g{\l(8)} \\
+\;&\b{\l(6)} &&- \r{\l(7)} &&- \g{\l(8)} &&+ {\l(9)} \\
+\; &\cdots
\end{alignat*}
Notice how each diagonal always has the same numbers. In particular, once we get to full diagonals of length $4$, there's always two $+\l(x)$ terms and two $-\l(x)$ terms, meaning that diagonal sums, ultimately, to zero.
Hence, the only terms left "not cancelled" in the infinite limit of your sum are those of the first three diagonals.

*

*$\l(1) = 0$ trivially.

*The two $\l(2)$ terms negate.

*Of the three $\l(3)$ terms, only a $-\l(3)$ term will remain.

Thus,
$$\lim_{n \to \infty} \sum_{k=1}^n \l \p{ 1-\frac{2}{k(k+1)} } = -\l(3) = 1$$
on the assumption the sequence of partial sums converges.
For the sequence of partial sums to converge, though, note that we eventually have partial sums of the form
$$
-\l(3) - \l(k) + \l(k+2)
$$
for $k \ge 4$ being the $k$th partial sum. We want this to converge as $k\to\infty$. Of course,
\begin{align*}
-\l(3) - \l(k) + \l(k+2) 
&= -\l(3) - \l\p{\frac{k+2}{k}}\\
&= -\l(3) - \underbrace{\l\p{1 + \frac 2 k}}_{\overset{k\to\infty}{\longrightarrow}\l(1)=0}\\
\end{align*}
giving the desired result, so everything's alright.
