<s>[INST] 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?
 [/INST] Given $$1+\frac{1}{k^2}+\frac{1}{(k+1)^2} = 1+\frac{1}{k^2}+\frac{1}{(k+1)^2}-\frac{2}{k(k+1)}+\frac{2}{k(k+1)}$$
So $$1^2+\left[\frac{1}{k}-\frac{1}{(k+1)}\right]^2+2\left[\frac{1}{k}-\frac{1}{(k+1)}\right]=\left[1+\frac{1}{k}-\frac{1}{k+1}\right]^2$$
</s>