Chapter 12.4, Problem 3CFU

(a)

To determine

To sketch: the graph of the first eight partial sum of the series.

Answer to Problem 3CFU

The graph is increasing parabolic.

Explanation of Solution

Given:

  $\frac{1}{3}+\frac{2^2}{3^2}+\frac{3^2}{3^3}+\frac{4^2}{3^4}+\mathrm{......}$

Concept used:

A partial sum of an infinite series is the sum of a finite number of consecutive terms beginning with the first term, when working with the infinite series, it is examined the behaviour of the partial sums.

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Calculation:

Let’s consider the infinite series as:

  $\frac{1}{3}+\frac{2^2}{3^2}+\frac{3^2}{3^3}+\frac{4^2}{3^4}+\mathrm{......}$

Using spread sheet, the following tables show the first $8$ partial sums of the above series:

Number	Partial sum
$1$	$0.3333333$
$2$	$0.7777778$
$3$	$1.4444444$
$4$	$2.7530864$
$5$	$5.4115226$
$6$	$10.769547$
$7$	$21.512117$
$8$	$43.011584$

The graph of the first eight partial sum of the above series can be drawn with the help of the above table as below:

  

Hence, the graph is increasing parabolic.

(b)

To determine

To find: the conjecture based on the graph in a part a as series is convergent or divergent.

Answer to Problem 3CFU

It is a convergent.

Explanation of Solution

Given:

  $\frac{1}{3}+\frac{2^2}{3^2}+\frac{3^2}{3^3}+\frac{4^2}{3^4}+\mathrm{......}$

Concept used:

Calculation:

If an infinite series has a sum or limit, the series is convergent.

By observing the partial sums of the series, the series is convergent.

Hence, it is a convergent.

(c)

To determine

To find: the general term for the series.

Answer to Problem 3CFU

  $t_n=\frac{n^2}{3^n}$ .

Explanation of Solution

Given:

  $\frac{1}{3}+\frac{2^2}{3^2}+\frac{3^2}{3^3}+\frac{4^2}{3^4}+\mathrm{......}$

Concept used:

A partial sum of an infinite series is the sum of a finite number of consecutive terms beginning with the first term, when working with the infinite series, it is examined the behaviour of the partial sums.

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Calculation:

The general term of the series is:

  $t_n=\frac{n^2}{3^n}$ .

Hence, general formula for nth term is $t_n=\frac{n^2}{3^n}$ .

(d)

To determine

To find: the convincing argument that support the conjecture made in part b.

Answer to Problem 3CFU

It is clear the series is convergent.

Explanation of Solution

Given:

  $\frac{1}{3}+\frac{2^2}{3^2}+\frac{3^2}{3^3}+\frac{4^2}{3^4}+\mathrm{......}$

Concept used:

Ratio test:

If the value of $\underset{n\to \infty }{\lim }\frac{a_n+1}{a_n}$ is less than $1$ it is convergent , if that value is more than $1$ then the series is divergent.

Calculation:

By ratio test, it can check the above series is convergent or not.

By ratio test if the value of $\underset{n\to \infty }{\lim }\frac{a_n+1}{a_n}$ is less than $1$ it is convergent , if that value is more than $1$ then the series is divergent.

Here,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{a_n+1}{a_n}=\underset{n\to \infty }{\lim }\frac{\frac{{\left(n+1\right)}^2}{3^{n+1}}}{\frac{n^2}{3^n}}.\\ \Rightarrow \underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^2\frac{1}{3}.\\ \Rightarrow \frac{1+0}{3}=\frac{1}{3}<1.\end{array}$

Hence, it is clear the series is convergent.