Telescoping series are series where we can find their sum by canceling out terms.
$$ \sum_{n=2}^\infty\ln\left(\frac{n+1}{n-1}\right) $$
To find this sum, we can use logarithm properties to rewrite the sum:
$$ \sum_{n=2}^\infty\ln\left(\frac{n+1}{n-1}\right) = \sum_{n=2}^\infty[\ln(n+1)-\ln(n-1)] $$
Now let’s try writing this out term-by-term:
$$ \sum_{n=2}^\infty[\ln(n+1)-\ln(n-1)] = [\ln(3)-\ln(1)] + [\ln(4)-\ln(2)] + [\ln(5)-\ln(3)] + \cdots $$
Notice how the first $\ln(3)$ term is followed by a $-\ln(3)$ term later on, and the same can be said about the $\ln(4)$ term, the $\ln(5)$ term, etc. So it seems like we can cancel most of the terms:
$$ \sum_{n=2}^\infty[\ln(n+1)-\ln(n-1)] = [\cancel{\ln(3)}-\ln(1)] + [\cancel{\ln(4)}-\ln(2)] + [\cancel{\ln(5)}\cancel{-\ln(3)}] + \cdots $$
However, we can’t quite conclude yet that the sum is $-\ln(1) -\ln(2)$.
Remember that an infinite sum is really just a limit of partial sums. By using a similar argument involving canceling terms, we can say that the finite partial sum is equal to:
$$ \sum_{n=2}^N[\ln(n+1)-\ln(n-1)] = \ln(N+1) + \ln(N+2) - \ln(1) - \ln(2) $$
(Technically, this is only true for $N \ge 3$.)
To find the sum of the infinite series, we need to take the limit of this expression as $N$ approaches infinity. However, if we do that, we find that the series actually diverges!
$$ \sum_{n=2}^\infty[\ln(n+1)-\ln(n-1)] = \lim_{n\to\infty}\sum_{n=2}^N[\ln(n+1)-\ln(n-1)] = \lim_{n\to\infty}[\ln(N+1)+\ln(N+2)-\ln(1)-\ln(2)] = \infty $$
$$ \sum_{n=1}^\infty\frac{2}{4n^2-1} = \sum_{n=1}^\infty\frac{2}{(2n+1)(2n-1)} = \sum_{n=1}^\infty\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) $$
(I did this using partial fraction decomposition - an explanation of that can be found here.)
We can find that the $N$th partial sum is:
$$ \sum_{n=1}^N\left[\frac{1}{2n-1} - \frac{1}{2n+1}\right] = 1-\frac{1}{2N+1} $$