## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty}\left(\frac{n+2}{1+2+3+\ldots+n}-\frac{2}{3}\right)
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+2}{1+2+3+\ldots+n}-\frac{2}{3}\right)=\lim _{n \rightarrow \infty}\left(\frac{n+2}{\frac{(1+n) n}{2}}-\frac{2}{3}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{2 n+4}{n+n^{2}}-\frac{2}{3}\right)=\lim _{n \rightarrow \infty} \frac{2 n+4}{n+n^{2}}-\lim _{n \rightarrow \infty} \frac{2}{3}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}(2 n+4)}{\frac{1}{n^{2}}\left(n+n^{2}\right)}-\frac{2}{3}=\lim _{n \rightarrow \infty} \frac{\frac{2}{n}+\frac{4}{n^{2}}}{\frac{1}{n}+1}-\frac{2}{3}=
\end{aligned}
$$

$=\frac{0+0}{0+1}-\frac{2}{3}=-\frac{2}{3}$

## Problem Kuznetsov Limits 6-17