Summation of a logarithmic series for $\ln(2(r^2 - 1)/r^2)$

Question

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 $A$.

Answer 1

You don't need to deal with huge numbers.$$\begin{align}\sum_{r=32}^{62}\ln\left(\frac{2(r^2-1)}{r^2}\right)&=\sum_{r=32}^{62}\left(\ln 2+\ln\left(\frac{r^2-1}{r^2}\right)\right)\\&=\sum_{r=32}^{62}\ln 2+\sum_{r=32}^{62}\ln\left(\frac{r^2-1}{r^2}\right)\\&=(62-32+1)\ln 2+\sum_{r=2}^{62}\ln\left(\frac{r^2-1}{r^2}\right)-\sum_{r=2}^{31}\ln\left(\frac{r^2-1}{r^2}\right)\\&=31\ln 2+\ln\frac{62+1}{2\cdot 62}-\ln\frac{31+1}{2\cdot 31}\\&=31\ln 2+\ln(3^2\cdot 7)-\ln(2^2\cdot 31)-\ln(2^5)+\ln(2\cdot 31)\\&=31\ln 2+2\ln 3+\ln 7-2\ln 2-\ln(31)-5\ln 2+\ln 2+\ln(31)\\&=25\ln 2+2\ln 3+\ln 7\end{align}$$

Answer 2

If one was unaware of the identity, one could observe that

$$\log\left(\frac{r^2-1}{r^2}\right)=\left(\log (r+1)-\log r\right)-\left(\log r-\log (r-1)\right)$$

from which the sum of interest is decomposed into two telescoping sums. Thus, we have

$$\begin{align} \sum_{r=m}^n \log\left(2\frac{r^2-1}{r^2}\right)&=(n-m+1)\log 2+\sum_{r=m}^n \left(\log (r+1)-\log r\right)-\sum_{r=m}^n \left(\log r-\log (r-1)\right)\\\\ &=(n-m+1)\log 2+\left(\log(n+1)-\log(m)\right)-\left(\log (n)-\log(m-1)\right) \end{align}$$

from which the general result is expressed as

$$\bbox[5px,border:2px solid #C0A000]{\sum_{r=m}^n \log\left(2\frac{r^2-1}{r^2}\right)=(n-m+1)\log 2+\log\left(\frac{(n+1)(m-1)}{nm}\right)}\tag 1$$

For $m=2$, we have

$$\bbox[5px,border:2px solid #C0A000]{\sum_{r=2}^n \log\left(2\frac{r^2-1}{r^2}\right)=(n-1)\log 2+\log \frac{n+1}{2n}}$$

as expected.

And finally, using $(1)$ with $m=32$ and $n=62$ reveals

$$\bbox[5px,border:2px solid #C0A000]{\sum_{r=32}^{62} \log\left(2\frac{r^2-1}{r^2}\right)=31\log 2+\log\frac{(63)(31)}{(62)(32)} =\log (7)+2\log (3)+25\log 2}$$