<s>[INST] Summation of a logarithmic series for $\ln(2(r^2 - 1)/r^2)$ Given that
$$\sum_{r=2}^{n}\ln\frac{r^2-1}{r^2}=\ln\frac{n+1}{2n}$$
for $n >1$.
Express $$\sum_{r=32}^{62}{\ln\frac{2(r^2-1)}{r^2}}$$ as $$A\ln 2 + B\ln3 + C\ln7$$ where $A$, $B$, $C$ are positive integers that can be found.
The result is actually $\ln 2113929216$, which is $2^{25}$ multiplied by $3^2 \times 7 (=63)$, hence the $\ln 2$, $\ln3$ and $\ln 7$.
How would you show this without knowing the result? 
The upper limit is $62$, which is one away from the $3^2 \times 7$ for the $63$, leaving $2^{25}$ for the first term, with $A$.
 [/INST] You don't need to deal with huge numbers.$$\begin{align}\sum_{r=32}^{62}\ln\left(\frac{2(r^2-1)}{r^2}\right)&=\sum_{r=32}^{62}\left(\ln 2+\ln\left(\frac{r^2-1}{r^2}\right)\right)\\&=\sum_{r=32}^{62}\ln 2+\sum_{r=32}^{62}\ln\left(\frac{r^2-1}{r^2}\right)\\&=(62-32+1)\ln 2+\sum_{r=2}^{62}\ln\left(\frac{r^2-1}{r^2}\right)-\sum_{r=2}^{31}\ln\left(\frac{r^2-1}{r^2}\right)\\&=31\ln 2+\ln\frac{62+1}{2\cdot 62}-\ln\frac{31+1}{2\cdot 31}\\&=31\ln 2+\ln(3^2\cdot 7)-\ln(2^2\cdot 31)-\ln(2^5)+\ln(2\cdot 31)\\&=31\ln 2+2\ln 3+\ln 7-2\ln 2-\ln(31)-5\ln 2+\ln 2+\ln(31)\\&=25\ln 2+2\ln 3+\ln 7\end{align}$$
</s>