Answer: \(\tan^{-1}(x+1)-\tan^{-1}(x-1)=\tan^{-1}\left(\frac2{x^2}\right)\).

The sum to \(n\) terms is \(\tan^{-1}(n+1)+\tan^{-1}n-\frac\pi4\).

Let \(A=\tan^{-1}(x+1)\) and \(B=\tan^{-1}(x-1)\). Then \(\tan A=x+1\) and \(\tan B=x-1\).

Using \(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\),

\(\tan(A-B)=\frac{(x+1)-(x-1)}{1+(x+1)(x-1)}=\frac2{x^2}\).

Therefore

\(\tan^{-1}(x+1)-\tan^{-1}(x-1)=\tan^{-1}\left(\frac2{x^2}\right)\).

Now put \(x=1,2,3,\ldots,n\). For \(x=k\),

\(\tan^{-1}\left(\frac2{k^2}\right)=\tan^{-1}(k+1)-\tan^{-1}(k-1)\).

Thus the sum to \(n\) terms is

\(\sum_{k=1}^{n}\tan^{-1}\left(\frac2{k^2}\right)=\sum_{k=1}^{n}\left[\tan^{-1}(k+1)-\tan^{-1}(k-1)\right]\).

Writing out the terms gives

\((\tan^{-1}2-\tan^{-1}0)+(\tan^{-1}3-\tan^{-1}1)+\cdots+(\tan^{-1}n-\tan^{-1}(n-2))+(\tan^{-1}(n+1)-\tan^{-1}(n-1))\).

The middle terms cancel, leaving

\(\tan^{-1}(n+1)+\tan^{-1}n-\tan^{-1}1-\tan^{-1}0\).

Since \(\tan^{-1}1=\frac\pi4\) and \(\tan^{-1}0=0\), the sum is

\(\tan^{-1}(n+1)+\tan^{-1}n-\frac\pi4\).