### Instruction: What is the value of $ \sum_{n=1}^\infty  (\tan^{-1}\sqrt{n}-\tan^{-1}\sqrt{n+1})$?

Your answer should be in radians.### Response: I recognize this as a telescoping series, where each term cancels out with another term except for the first and the last ones.
To see this more clearly, I can write out the first few terms of the series:

$ \sum_{n=1}^\infty  (\tan^{-1}\sqrt{n}-\tan^{-1}\sqrt{n+1})=\tan^{-1}\sqrt{1}-\tan^{-1}\sqrt{2}+\tan^{-1}\sqrt{2}-\tan^{-1}\sqrt{3}+\tan^{-1}\sqrt{3}-\tan^{-1}\sqrt{4}+...$
As n goes to infinity, the last term $\tan^{-1}\sqrt{n+1}$ goes to $\frac{\pi}{2}$, since the inverse tangent function approaches that value as its argument goes to infinity.
Therefore, the series simplifies to:

$ \sum_{n=1}^\infty  (\tan^{-1}\sqrt{n}-\tan^{-1}\sqrt{n+1})=\tan^{-1}\sqrt{1}-\frac{\pi}{2}$
I know that $\tan^{-1}\sqrt{1}=\frac{\pi}{4}$, since the tangent function is equal to 1 at that angle.