Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 5 - The Integral - 5.1 Approximating and Computing Area - Exercises - Page 236: 44

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}

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