The sum telescopes. Consider the N-th partial sum of the series,
[tex]S_N=\displaystyle\sum_{n=1}^N\bigg(\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n+1}\bigg)[/tex]
[tex]S_N=\bigg(\tan^{-1}\sqrt1-\tan^{-1}\sqrt2\bigg)+\bigg(\tan^{-1}\sqrt2-\tan^{-1}\sqrt3\bigg)+\bigg(\tan^{-1}\sqrt3-\tan^{-1}\sqrt4\bigg)+\cdots+\bigg(\tan^{-1}\sqrt N-\tan^{-1}\sqrt{N+1}\bigg)[/tex]
Notice how [tex]\tan^{-1}\sqrt2[/tex] is added, then immediately subtracted. The same goes for [tex]\tan^{-1}\sqrt3[/tex], then [tex]\tan^{-1}\sqrt4[/tex], and so on, up to [tex]\tan^{-1}\sqrt N[/tex]. This leaves us with
[tex]S_N=\tan^{-1}\sqrt1-\tan^{-1}\sqrt{N+1}[/tex]
The value of the given sum is obtained as [tex]N\to\infty[/tex]. We get
[tex]\displaystyle\sum_{n=1}^\infty\bigg(\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n+1}\bigg)=\lim_{N\to\infty}S_N=\tan^{-1}1-\lim_{N\to\infty}\tan^{-1}\sqrt{N+1}[/tex]
To compute the limit, recall that [tex]\tan^{-1}x[/tex] has a range of [tex]\left(-\frac\pi2,\frac\pi2\right)[/tex]; in particular, [tex]\tan^{-1}x\to\frac\pi2[/tex] as [tex]x\to\infty[/tex]. So we have
[tex]\displaystyle\sum_{n=1}^\infty\bigg(\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n+1}\bigg)=\frac\pi4-\frac\pi2=\boxed{-\frac\pi4}[/tex]
