Q1Advanced Engineering Mathematics I

Given $f(n)={\tan }^{-1}\left(\frac{n-1}{n}\right)$.

$\Delta f(n)=f(n+1)-f(n)={\tan }^{-1}\left(\frac{n}{n+1}\right)-{\tan }^{-1}\left(\frac{n-1}{n}\right)$.

Using the trigonometric identity ${\tan }^{-1}(a)-{\tan }^{-1}(b)={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$:

$={\tan }^{-1}\left(\frac{\frac{n}{n+1}-\frac{n-1}{n}}{1+\frac{n(n-1)}{n(n+1)}}\right)$

Simplifying the algebraic expression inside the arctangent yields ${\tan }^{-1}\left(\frac{1}{2n^2}\right)$. Proved.