<s>[INST] How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$ How to find the sum of the following : 
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
Please suggest as getting no clue on this... thanks..
 [/INST] Hint:
Let us write it as follows:
$$\sum_{n=1}^{1999} \sqrt{1+ \frac{1}{n^{2}} + \frac{1}{(n+1)^{2}}}$$
We can rewrite the radical as:
$$\sqrt{\frac{n^{2}(n+1)^{2} + n^{2} + (n+1)^{2}}{n^{2}(n+1)^{2}}}$$
Which simplifies to:
$$\sqrt{\frac{(n^{2}+n+1)^{2}}{n^{2}(n+1)^{2}}} = \frac{n^{2}+n+1}{n^{2}+n} = 1 + \frac{1}{n^{2}+n}$$
So now our sum is:
$$\sum_{n=1}^{1999} 1 + \frac{1}{n^{2}+n}$$
Extracting the constant term gives
$$\sum_{n=1}^{1999} 1 + \frac{1}{n^{2}+n} = 1999 + \sum_{n=1}^{1999}\frac{1}{n(n+1)}$$
By partial fractions decomposition, we see:
$$\frac{1}{n(n+1)} = \frac{1}{n} -\frac{1}{n+1}$$
The series therefore telescopes, making the sum easy to compute.
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