Find the value of $$\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \sqrt{1 + \frac{1}{3^2} + \frac{1}{4^2}} +...+ \sqrt{1 + \frac{1}{2010^2} + \frac{1}{2011^2}}$$

I have not been able to simplify the terms. Although there is a clear pattern, it does not result in any easy solution. This was a MCQ type question and the options were really simple terms, so I am assuming that this can be simplified really neatly. Please help.

  • 1
    $\begingroup$ Not linear-algebra. $\endgroup$
    – Calvin Lin
    Commented May 10, 2014 at 14:04

1 Answer 1


Let us first note that $$n^2+(n-1)^2+n^2(n-1)^2=(n^2-n+1)^2,$$ and therefore the sum can be rewritten as $$\sum_{n=2}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n-1)^2}}=\sum_{n=2}^{2011}\frac{n^2-n+1}{n(n-1)}=\sum_{n=2}^{2011}\left(1+\frac{1}{n-1}-\frac1n\right)=2011-\frac{1}{2011}.$$ The sum of $1$'s at the last step gives $2010$. The other two terms produce a telescoping series so that only the first (equal to $1$) and last (equal to $-\frac{1}{2011}$) terms survive.


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