Chapter F, Problem 43E

To determine

To find: The value of $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}{\left(\frac{i}{n}\right)}^2$.

Answer to Problem 43E

The value of $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}{\left(\frac{i}{n}\right)}^2$ is $\frac{1}{3}$.

Explanation of Solution

Theorem used:

Let c be a constant and n be a positive integer. Then,

${\displaystyle \sum _{i=1}^{n}c}=nc$, ${\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$.

Calculation:

Simplify the expression $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}{\left(\frac{i}{n}\right)}^2$ and obtain the limit as follows.

$\begin{array}{c}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}{\left(\frac{i}{n}\right)}^2=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}\left(\frac{i^2}{n^2}\right)\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n^3}}\left(i^3\right)\\ =\underset{n\to \infty }{\lim }\frac{1}{n^3}{\displaystyle \sum _{i=1}^n{i}^3}\\ =\underset{n\to \infty }{\lim }\frac{1}{n^3}\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\end{array}$

On further simplification the value of the limit becomes,

$\begin{array}{c}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}{\left(\frac{i}{n}\right)}^2=\underset{n\to \infty }{\lim }\frac{1}{n^3}\left(\frac{n^2\left(1+\frac{1}{n}\right)n\left(2+\frac{1}{n}\right)}{6}\right)\\ =\underset{n\to \infty }{\lim }\frac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\\ =\frac{\left(1+0\right)\left(2+0\right)}{6}\left[\because \underset{n\to \infty }{\lim }\frac{1}{n}=0\right]\\ =\frac{1}{3}\end{array}$

Thus, value of $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n}}{\left(\frac{i}{n}\right)}^2$ is $\frac{1}{3}$.