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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 $a^2+b^2/(a-b)(a+b)$.

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  • $\begingroup$ do long division and partial fractions. you will get a telescoping series for the fractions $\endgroup$ – David Quinn Feb 15 '16 at 11:25
  • $\begingroup$ What do you mean with "lies in the interval"? $\endgroup$ – Bobson Dugnutt Feb 15 '16 at 11:25
  • $\begingroup$ Like there is some answer x and the book has given these options where it will lie. $\endgroup$ – Archis Welankar Feb 15 '16 at 11:26
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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 $(2011,2011\frac12)$.

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Hint: $$\frac{n^{2}+1}{n^{2}-1}=1+\frac{1}{n-1}-\frac{1}{n+1}. $$

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