## Task Condition

Calculate the limit of the numerical sequence:

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

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