Summation of this huge series The value of $\dfrac{2^2+1}{2^2-1}+\dfrac{3^2+1}{3^2-1}...+\dfrac{2011^2+1}{2011^2-1}$ is:


*

*In the interval $(2010,2010\frac{1}{2})$

*In the interval $(2011-1/2011,2011-1/2012)$

*In the interval $(2011,2011\frac{1}{2})$

*In the interval $(2012,2012\frac{1}{2})$


I'm staring at it but can't see any trick to solve it.
I think there is some trick as its of the form $a^2+b^2/(a-b)(a+b)$.

Note that $$\frac{k^2+1}{k^2-1}=\frac{k^2-1}{k^2-1}+\frac{2}{k^2-1}=1+\frac{2}{k^2-1}=1+\frac{1}{k-1}-\frac{1}{k+1}$$
So that \begin{align}
\sum_{k=2}^{2011}\frac{k^2+1}{k^2-1}&=\sum_{k=2}^{2011}\left(1+\frac{2}{k^2-1}\right)\\
&=2010+\sum_{k=2}^{2011}\frac{1}{k-1}-\sum_{k=2}^{2011}\frac{1}{k+1}\\
&=2010+\frac{1}{1}+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011}+\sum_{k=3}^{2010}\frac{1}{k}-\sum_{k=3}^{2010}\frac{1}{k}\\
&=2010+\frac{1}{1}+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011}
\end{align}
so that is just a little less than $2011\frac12$, so it's in the interval $(2011,2011\frac12)$.
