Answer
$$\frac{1}{4}$$
Work Step by Step
Given $$ \lim _{N \rightarrow \infty} \sum_{i=1}^{N} \frac{i}{N^{2}}$$
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{aligned}
\lim _{N \rightarrow \infty} \sum_{j=1}^{N} \frac{j^{3}}{N^{4}} &= \frac{1}{N^{4}}\lim _{N \rightarrow \infty} \sum_{j=1}^{N} j^{3},\ \ \ \ \text{use (3)} \\
&=\lim _{N \rightarrow \infty} \frac{1}{N^{4}}\left(\frac{N^{2}(N+1)^{2}}{4}\right) \\
&=\lim _{N \rightarrow \infty} \frac{\left(N^{2}+1+2 N\right)}{4 N^{2}}\\
&=\frac{1}{4}
\end{aligned}
