## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{\sqrt{9 n^{4}+1}}
$$
## Solution

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