Test for convergence for $\ln \frac{n^2}{n^2-1}$ I've tried to figure out if this converges using the comparison test, and the ratio test, but with no luck: $\sum^\infty_{n=2} \ln(n^2/(n^2-1))$.
I'd appreciate any help
 A: Using the properties of logarithms,
I write $\ln(n^2/(n^2-1))$ 
as 
$2\ln(n)-\ln(n-1)-\ln(n+1)$
and then I consider the partial sum $S_n= (2\ln(2)-\ln(1)-\ln(3))+(2\ln(3)-\ln(2)-\ln(4))+(2\ln(4)-\ln(3)-\ln(5))+...+(2\ln(n)-\ln(n-1)-\ln(n+1))$
Which after canceling like terms simplifies to
$S_n=\ln2+\ln(n)-\ln(n+1)=\ln(2n/(n+1))=\ln(2/(1+(1/n)))$
And since $\lim_{n\rightarrow\infty}(2/(1+1/n))=\ln(2).$ 
Since $\ln(2) \in \Re$, the series must converge to $\ln(2)$. 
A: Here is a way
to show convergence
without finding the sum.
If $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
$.
Therefore
(the $1+x$ in the denominator
of the next integral
is intentional)
$\int_1^{1+x} \frac{dt}{1+x}
< \ln(1+x)
< \int_1^{1+x} \frac{dt}{1}
$
or
$\frac{x}{1+x}
< \ln(1+x)
< x
$.
Since
$\frac{n^2}{n^2-1}
=\frac{n^2-1+1}{n^2-1}
=1+\frac{1}{n^2-1}
$,
$\ln \frac{n^2}{n^2-1}
<\frac{1}{n^2-1}
$.
Since
$\sum^\infty_{n=2} \frac{1}{n^2-1}
$ converges,
so does
$\sum^\infty_{n=2} \ln(n^2/(n^2-1))$.
For a lower bound,
$\ln \frac{n^2}{n^2-1}
>\frac{\frac{1}{n^2-1}}{1+\frac{1}{n^2-1}}
=\frac{1}{n^2}
$.
Note that,
in general,
if $m > 1$,
$\frac{1}{m}
<\ln \frac{m}{m-1}
<\frac{1}{m-1}
$.
