limit of sum

$\begin{aligned} L & = \lim_{n \to \infty} \sum_{i=1}^n \frac 3n \left( \left( \frac in \right)^2 + 1 \right) & \small \blue{\text{By Riemann sums}} \\ & = 3 \int_0^1 (x^2 + 1) \ dx \\ & = 3 \left[\frac {x^3}3+x \right]_0^1 \\ & = \boxed 4 \end{aligned}$

Reference: Riemann sums

The sum is $3\displaystyle \int_0^1 (x^2+1)dx=3(\frac{1}{3}+1)=\boxed 4$ .

The above summation computes to $\frac{6n^3 +9n^2 + 3n}{6n^3} + 3 = 1 + \frac{3}{2n} + \frac{1}{2n^2} + 3$ , which has a limit of $4$ as $n \rightarrow \infty.$