## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty}\left(\frac{1+5+9+13+\ldots+(4 n-3)}{n+1}-\frac{4 n+1}{2}\right)
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{1+5+9+13+\ldots+(4 n-3)}{n+1}-\frac{4 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{\frac{(1+(4 n-3)) n}{2}}{n+1}-\frac{4 n+1}{2}\right)=\lim _{n \rightarrow \infty}\left(\frac{(4 n-2) n}{2(n+1)}-\frac{4 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{(4 n-2) n}{2(n+1)}-\frac{(4 n+1)(n+1)}{2(n+1)}\right)=\lim _{n \rightarrow \infty}\left(\frac{(4 n-2) n-(4 n+1)(n+1)}{2(n+1)}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{4 n^{2}-2 n-4 n^{2}-n-4 n-1}{2(n+1)}=\lim _{n \rightarrow \infty} \frac{-7 n-1}{2(n+1)}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(-7 n-1)}{\frac{1}{n} 2(n+1)}=\lim _{n \rightarrow \infty} \frac{-7-\frac{1}{n}}{2\left(1+\frac{1}{n}\right)}=\frac{-7-0}{2(1+0)}=-\frac{7}{2}
\end{aligned}
$$

## Problem Kuznetsov Limits 6-25