Cfrgtkky

According to Wolfram Alpha,
$$sum_{n=1}^{infty}frac{1}{n(2n+1)(2n-1)}=ln(4)-1$$
However, I am not sure how to evaluate this series.

Attempt
$$frac{1}{n(2n+1)(2n-1)}=frac{A}{n}+frac{B}{2n+1}+frac{C}{2n-1}$$
$$1=A(2n+1)(2n-1)+Bn(2n-1)+Cn(2n+1)$$

Then, I got
$$
sum_{n=1}^{infty}frac{1}{n(2n+1)(2n-1)} = sum_{n=1}^{infty}frac{-1}{n}+frac{1}{2n+1}+frac{1}{2n-1}\
$$
I tried to view this as a telescoping series, but it did not turn out good. Can I have a hint?

$$
sum_{ngeq 1} left( frac{1}{2n-1} - frac{1}{2n}right) - sum_{ngeq 1} left( frac{1}{2n} - frac{1}{2n+1}right) = ln 2 - (1-ln 2)
$$

Too long for a comment.

You could even get a good approximation of the partial sums
$$S_p=sum_{n=1}^{p}frac{1}{n(2n+1)(2n-1)}=sum_{n=1}^{p}frac{1}{2 n-1}+sum_{n=1}^{p}frac{1}{2 n+1}-sum_{n=1}^{p}frac{1}{n}$$ that is to say
$$S_p=left(frac{1}{2}H_{p-frac{1}{2}}+log (2)right)+left(frac{1}{2}H_{p+frac{1}{2}}-1+log (2)right)-H_p$$

Now, using the asymptotics of harmonic numbers
$$H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
n^2}+frac{1}{120 n^4}+Oleft(frac{1}{n^6}right)$$ replacing and continuing with Taylor series, you could get
$$S_p=(2 log (2)-1)-frac{1}{8 p^2}+frac{1}{8 p^3}-frac{5}{64 p^4}+Oleft(frac{1}{p^5}right)$$

For $n=5$, the exact value is $S_5=frac{5297}{13860}approx 0.3821789$ while the above approximation would give $log (4)-frac{200823}{200000}approx 0.3821794$.