Chapter 8, Problem 14P

To determine

To prove: The series ${\displaystyle \sum _{n=1}^{\infty }{\theta }_n}$ is divergent.

Answer to Problem 14P

The proof is given below.

Explanation of Solution

Given:

The given figure is as:

  

Calculation:

Let length of the base of the first triangle be $a$ .

By the Pythagorean Theorem, the length of the hypotenuse of the first triangle is $\sqrt{1+a^2}$ .

Hypotenuse of the first triangle is the base of the second triangle.

So the base of the second triangle is $\sqrt{1+a^2}$ .

Similarly, the base of the third triangle is $\sqrt{2+a^2}$ .

Similarly the length of the base of the $n^{\text{th}}$ triangle is $\sqrt{n-1+a^2}$ .

The length of the hypotenuse of the $n^{\text{th}}$ triangle is $\sqrt{n+a^2}$ .

Now using trigonometric ratio in the $n^{\text{th}}$ triangle, we get:

  $\begin{array}{l}\tan {\theta }_n=\frac{1}{\sqrt{n-1+a^2}}\\ {\theta }_n={\tan }^{-1}\left(\frac{1}{\sqrt{n-1+a^2}}\right)\end{array}$

Further simplified as:

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }{\theta }_n}={\displaystyle \sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{\sqrt{n-1+a^2}}\right)}\\ {\displaystyle \sum _{n=1}^{\infty }{\theta }_n}={\displaystyle \sum _{n=0}^{\infty }{\tan }^{-1}\left(\frac{1}{\sqrt{n+a^2}}\right)}\end{array}$

Now show that series ${\displaystyle \sum _{n=1}^{\infty }{\theta }_n}$ is divergent.

Consider the series ${\displaystyle \sum _{n=1}^{\infty }{u}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n+a^2}}}$ .

  $\text{Let}{v}_n={\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}}$ .

Use limit comparison test:

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\frac{1}{\sqrt{n+a^2}}}{\frac{1}{\sqrt{n}}}=\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{\sqrt{n+a^2}}\\ =\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{\sqrt{n}\sqrt{1+\frac{a^2}{n}}}\\ =\frac{1}{\sqrt{1+0}}\\ =1\left(\text{Finite}\right)\end{array}$

Therefore both series are convergent or divergent.

Since using the $p$ -test, the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}}$ is divergent.

Therefore by using the limit comparison test the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n+a^2}}}$ is divergent.

Consider the two series, ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n+a^2}}}$ and ${\displaystyle \sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{\sqrt{n+a^2}}\right)}$ .

  $\text{Let }\underset{n\to \infty }{\lim }\frac{{\tan }^{-1}\left(\frac{1}{\sqrt{n+a^2}}\right)}{\frac{1}{\sqrt{n+a^2}}}=l$

If $l$ is finite, both series are either convergent or divergent.

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{{\tan }^{-1}\left(\frac{1}{\sqrt{n+a^2}}\right)}{\frac{1}{\sqrt{n+a^2}}}=\underset{x\to \infty }{\lim }\frac{{\tan }^{-1}\left(\frac{1}{\sqrt{x+a^2}}\right)}{\frac{1}{\sqrt{x+a^2}}}\\ =\underset{y\to \infty }{\lim }\frac{{\tan }^{-1}\left(\frac{1}{y}\right)}{\frac{1}{y}}\left(\text{Let}y=\sqrt{x+a^2}\right)\\ =\underset{y\to \infty }{\lim }\frac{{\tan }^{-1}\left(z\right)}{z}\left(\text{Let}\frac{1}{y}=z\right)\\ =1\left(\text{Finite}\right)\end{array}$

Therefore, both are convergent or divergent.

But the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n+a^2}}}$ is divergent.

Therefore the series ${\displaystyle \sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{\sqrt{n+a^2}}\right)}$ is divergent.

Hence ${\displaystyle \sum _{n=1}^{\infty }{\theta }_n}$ is divergent.