## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty} \frac{3+6+9+\ldots+3 n}{n^{2}+4}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{3+6+9+\ldots+3 n}{n^{2}+4}=\lim _{n \rightarrow \infty} \frac{\left(\frac{(3+3 n) n}{2}\right)}{n^{2}+4}= \\
& =\lim _{n \rightarrow \infty} \frac{(3+3 n) n}{2\left(n^{2}+4\right)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}(3+3 n) n}{\frac{1}{n^{2}} 2\left(n^{2}+4\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{3}{n}+3}{2\left(1+\frac{4}{n^{2}}\right)}=\frac{0+3}{2(1+0)}=\frac{3}{2}
\end{aligned}
$$

## Problem Kuznetsov Limits 6-29