Chapter 8.3, Problem 1E

To determine

To prove: ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^{1.3}}<{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{1.3}}}dx}$ by drawing the pictures and to conclude about the series.

Answer to Problem 1E

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^{1.3}}}$ is convergent.

Explanation of Solution

Given Information: Pictures must be drawn to prove the result.

Graph:

  

Proof:

The sum of the areas of the rectangles represents the given series.

From the figure, the total area of the rectangles is less than the area under the curve $y=\frac{1}{x^{1.3}}$ for $x\ge 1$ which is the value of ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{1.3}}}dx$ .

Hence, ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^{1.3}}<{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{1.3}}}dx}$ .

Now, the integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{1.3}}}dx$ converges, so by the Integral Test, ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^{1.3}}}$ also converges.