## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty}\left(\frac{1+3+5+7+\ldots+(2 n-1)}{n+1}-\frac{2 n+1}{2}\right)
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{1+3+5+7+\ldots+(2 n-1)}{n+1}-\frac{2 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{n+1} \cdot \frac{(1+(2 n-1)) n}{2}-\frac{2 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{n+1} \cdot \frac{2 n^{2}}{2}-\frac{2 n+1}{2}\right)=\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}}{2(n+1)}-\frac{(2 n+1)(n+1)}{2(n+1)}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{2 n^{2}-(2 n+1)(n+1)}{2(n+1)}=\lim _{n \rightarrow \infty} \frac{2 n^{2}-\left(2 n^{2}+2 n+n+1\right)}{2(n+1)}= \\
& =\lim _{n \rightarrow \infty} \frac{2 n^{2}-2 n^{2}-2 n-n-1}{2(n+1)}=\lim _{n \rightarrow \infty} \frac{-3 n-1}{2(n+1)}=
\end{aligned}
$$

$$
=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(-3 n-1)}{\frac{1}{n} 2(n+1)}=\lim _{n \rightarrow \infty} \frac{-3-\frac{1}{n}}{2+\frac{2}{n}}=\frac{-3-0}{2+0}=-\frac{3}{2}
$$

## Problem Kuznetsov Limits 6-3