On simplifying the given series, we get:
\((1-\frac{1}{n})+(1-\frac{2}{n})+(1-\frac{3}{n})\)+.....n terms

Here, \((\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+....+\frac{n}{n})\) is an AP whose first term is \(\frac{1}{n}\)
and the common difference is \((\frac{2}{n}-\frac{1}{n})=\frac{1}{n}.\)
The sum of terms of an AP is given by
