Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$ Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
I found the general term of the sequence.
It is $\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
So the sequence becomes $\sum_{k=1}^{1999}\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
I tried telescoping but i could not split it into two partial fractions.And this raised to $\frac{1}{2}$ is also troubling me.What should i do to find the answer?

After taking LCM, we get the general term of the series as:
$$\sqrt{\frac{(k^{2}+k+1)^{2}}{k^{2}(k+1)^{2}}} $$
$$=> \frac{k^{2}+k+1}{k^{2}+k}$$ 
$$=> 1 + \frac{1}{k^{2}+k}$$
So we have 
$$\sum_{k=1}^{1999} 1 + \frac{1}{k^{2}+k}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k(k+1)}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k} - \frac{1}{(k+1)}$$ 
$$=> 1999 + 1 - \frac{1}{2000}$$ 
$$=> 2000 - \frac{1}{2000}$$
