Experimentally (using Wolframalpha) I noticed that for $n=2,3,4,5$
$\log[n^2] - \log[n^2 - 1] = \sum((\frac{1}{n^4})^{k+1}\left(\frac {n^2}{(2k+1)}+\frac {1}{2k+2}\right),k = 0\cdots \infty$
Wolframalpha also confirms the generalization of the above as follows:
for $|n| > 1$
$\log[n^2] - \log[n^2 - 1] = \sum((\frac{1}{n^4})^{k+1}\left(\frac {n^2}{(2k+1)}+\frac {1}{2k+2}\right),k = 0\cdots \infty$
Could someone provide analytical prove for my above described generalization using induction?
A standard Maclaurin series is $$-\log(1-x)=\sum_{m=1}^\infty \frac{x^m}{m}\quad\hbox{for $|x|<1$}.$$ If $|n|>1$ then $0<1/n^2<1$ so we can substitute $x=1/n^2$ to get $$-\log\Bigl(1-\frac1{n^2}\Bigr)=\sum_{m=1}^\infty \frac1m\,\frac{1}{n^{2m}}\ .$$ Now sum the odd and even terms separately (we can do this because it's a convergent series of positive terms): take $m=2k+1$ for the odd terms, $m=2k+2$ for the even terms, both with $k$ from $0$ to $\infty$, giving $$\eqalign{-\log\Bigl(1-\frac1{n^2}\Bigr) &=\sum_{k=0}^\infty\Bigl(\frac1{2k+1}\,\frac1{n^{4k+2}}+\frac1{2k+2} \,\frac1{n^{4k+4}}\Bigr)\cr &=\sum_{k=0}^\infty\frac1{n^{4k+4}}\Bigl(\frac{n^2}{2k+1}+\frac1{2k+2}\Bigr)\ .\cr}$$ Since $$\log(n^2)-\log(n^2-1)=-\log\Bigl(\frac{n^2-1}{n^2}\Bigr)=-\log\Bigl(1-\frac1{n^2}\Bigr)\ ,$$ this is what you wanted.
