Answer
$$\frac{1}{3}$$
Work Step by Step
Given $$\lim _{N \rightarrow \infty} \sum_{i=1}^{N} \frac{i^{2}-i+1}{N^{3}} $$ Since \begin{align} \sum_{j=1}^{N} j&=1+2+\cdots+N=\frac{N(N+1)}{2} \tag{1}\\ \sum_{j=1}^{N} j^{2}&=1^{2}+2^{2}+\cdots+N^{2}=\frac{N(N+1)(2 N+1)}{6} \tag{2}\\ \sum_{j=1}^{N} j^{3}&=1^{3}+2^{3}+\cdots+N^{3}=\frac{N^{2}(N+1)^{2}}{4} \tag{3} \end{align} Then \begin{align*} \lim _{N \rightarrow \infty} \sum_{i=1}^{N} \frac{i^{2}-i+1}{N^{3}}&=\lim _{N \rightarrow \infty} \frac{1}{N^{3}} \sum_{i=1}^{N}\left(i^{2}-i+1\right)\\ &=\lim _{N \rightarrow \infty} \frac{1}{N^{3}} \left(\sum_{i=1}^{N}i^{2}-\sum_{i=1}^{N}i+\sum_{i=1}^{N}1\right),\ \ \text{Use (1),(2)}\\ &= \lim _{N \rightarrow \infty} \frac{1}{N^{3}}\left(\frac{N(N+1)(2 N+1)}{6}-\frac{N(N+1)}{2}+N\right)\\ &=\lim _{N \rightarrow \infty} \frac{1}{N^{3}}\left(\frac{N^{3}}{3}+\frac{N^{2}}{2}+\frac{N}{6}-\frac{N^{2}}{2}-\frac{N}{2}+N\right)\\ &=-\lim _{N \rightarrow \infty} \frac{1}{3}+\frac{1}{2 N}+\frac{1}{6 N^{2}}-\frac{1}{2 N}-\frac{1}{2 N^{2}}+\frac{1}{N^{2}}\\ &=\frac{1}{3} \end{align*}
