Demonstrate the telescoping nature of each series. Find the sum of the series if it converges for each series by including the general term Sn in its list of partial sums.

The series is$\sum _{k=2}^{\infty }\ln \frac{1}{k^2-1}$.

The goal is to demonstrate the series' telescopic nature, present the general term S {n} in its list of partial sums, and determine whether the series is convergent by determining its sum.

If each term of the series can be expressed as the sum of two or more summands and there is cancellation between terms in the sequence of partial sums, the series is telescopic.

The series $\sum _{k=2}^{\infty }\ln \frac{1}{k^2-1}$can be expressed as,

$\begin{array}{rcl}\sum _{k=2}^{\infty }\ln \frac{1}{k^2-1}& =& \sum _{k=2}^{\infty }\left[\ln \frac{1}{(k-1)(k+1)}\right]\\ & =& \sum _{k=2}^{\infty }[\ln 1-\ln ((k-1)(k+1)\left)\right]\left(\text{Because}\ln \frac{a}{b}=\ln a-\ln b\right)\\ & =& \sum _{k=2}^{\infty }[0-(\ln (k-1)+\ln (k+1)\left)\right]\\ & =& \sum _{k=2}^{\infty }[-\ln (k-1)-\ln (k+1\left)\right]\\ & =& (-\ln 1-\ln 3)+(-\ln 2-\ln 4)+\dots \end{array}$

In the series of partial sums, the nth term is defined as,

$\sum _{k=2}^{\infty }\ln \frac{1}{k^2-1}=(-\ln 1-\ln 3)+(-\ln 2-\ln 4)+\dots +(-\ln (n-1)-\ln (n+1\left)\right)$

The second term of a pair does not cancel with the first term of the following pair in any pair of two successive pairs.

The series is not telescopic as a result.

The generic phrase for the sum of the nth partial sums cannot be written since the series is not telescopic.

Therefore, in its series of partial sums, there is no general term S_n.

Since its series of partial sums lacks the general term S_n.