### 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: This looks like a telescoping series, where the terms cancel out except for a few at the beginning and end.
To see this more clearly, let me 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})+\dots$
Now I can see that each term except the first one has a negative part that cancels with the positive part of the next term.
For example, $-\tan^{-1}\sqrt{2}$ cancels with $+\tan^{-1}\sqrt{2}$.
So the only terms that do not cancel are the first positive part and the last negative part, which are $\tan^{-1}\sqrt{1}$ and $-\tan^{-1}\sqrt{\infty}$ respectively.
I can simplify these terms using some properties of the inverse tangent function.
First, $\tan^{-1}\sqrt{1}$ is the same as $\tan^{-1}1$, which is $\frac{\pi}{4}$ by definition.
Second, $\tan^{-1}\sqrt{\infty}$ is the same as $\lim_{x\to\infty}\tan^{-1}\sqrt{x}$, which is $\frac{\pi}{2}$ by the horizontal asymptote of the inverse tangent function.
So the value of the series is $\frac{\pi}{4}-\frac{\pi}{2}$, which is $-\frac{\pi}{4}$.