Answer
$$\frac{-79}{4} $$
Work Step by Step
Given $$\lim _{N \rightarrow \infty} \sum_{i=1}^{N}\left(\frac{i^{3}}{N^{4}}-\frac{20}{N}\right)$$ 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}\left(\frac{i^{3}}{N^{4}}-\frac{20}{N}\right)&=\lim _{N \rightarrow \infty} \frac{1}{N^{4}} \sum_{i=1}^{N} i^{3}-\lim _{N \rightarrow \infty} \frac{1}{N} \sum_{i=1}^{N} 20,\ \ \text{Use (3)}\\ &=\lim _{N \rightarrow \infty} \frac{1}{N^{4}} \frac{N^{2}(N+1)^{2}}{4}-\lim _{N \rightarrow \infty} \frac{1}{N} 20N\\ &=\frac{1}{4}-20=\frac{-79}{4} \end{align*}
