Question

The value of $\underset{n\to \infty }{\lim }\left\{\frac{1+2+3+...+n}{n+2}-\frac{n}{2}\right\}$ is

(a) 1/2

(b) 1

(c) −1

(d) −1/2

Answer 1

(d) −1/2

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{1+2+3+.....n}{n+2}-\frac{n}{2}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{n\left(n+1\right)}{2\left(n+2\right)}-\frac{n}{2}\right] \\ =\underset{n\to \infty }{\lim }\frac{n}{2}\left[\frac{n+1-n-2}{n+2}\right] \\ =\underset{n\to \infty }{\lim }\frac{n}{2}\left(\frac{-1}{n+2}\right) \\ =\underset{n\to \infty }{\lim }\frac{-1}{2\left(1+{\displaystyle \frac{2}{n}}\right)} \\ =\frac{-1}{2\left(1+0\right)} \\ =-\frac{1}{2} \end{gathered}$$