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

  • $\begingroup$ 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

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 $(2011,2011\frac12)$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.