Answer
$$\frac{1}{2}$$
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_{i=1}^{N} \frac{i}{N^{2}} &= \frac{1}{N^{2}}\lim _{N \rightarrow \infty} \sum_{i=1}^{N}i,\ \ \text{Use (1)}\\
&=\lim _{N \rightarrow \infty} \frac{1}{N^{2}}\left(\frac{N(N+1)}{2}\right) \\
&=\lim _{N \rightarrow \infty} \frac{(N+1)}{2 N}\\
&=\frac{1}{2}
\end{aligned}
