Chapter 12.9, Problem 28E

a.

To determine

Write $0.\overline{9}$ as an infinite series.

Answer to Problem 28E

  $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\mathrm{....}$

Explanation of Solution

Given information:

  $\mathrm{0.99999....}=1$

Calculation:

Here,

  $\begin{array}{l}\mathrm{0.9999999...}=\\ 0.9+0.09+0.009+0.0009+\mathrm{....}=\\ \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\mathrm{....}\end{array}$

Hence, required answer is $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\mathrm{....}$ .

b.

To determine

Write an expression for the $n^{th}$ term of the series.

Answer to Problem 28E

  $\frac{9}{{10}^n}$

Explanation of Solution

Given information:

  $\mathrm{0.99999....}=1$

Calculation:

Here, from above we can conclude that:

  $\frac{9}{{10}^n}$

Hence, the required answer is $\frac{9}{{10}^n}$.

c.

To determine

Write a formula for the sum of the first $n$ terms of the series.

Answer to Problem 28E

  $S_n=\frac{{10}^n-1}{{10}^n}$

Explanation of Solution

Given information:

  $\mathrm{0.99999....}=1$

Calculation:

Here, the ratios between neighbouring terms is constant.

  $\begin{array}{l}\frac{a_n+1}{a_n}=\frac{\frac{9}{{10}^{n+1}}}{\frac{9}{{10}^n}}=\frac{1}{10}=r\\ Thesumofthefirstnterms:\\ S_n=\frac{a_1-a_1{r}^n}{1-r}\\ S_n=\frac{\frac{9}{10}-\frac{9}{10}{\left(\frac{1}{10}\right)}^n}{1-\frac{1}{10}}\\ S_n=\frac{\frac{9}{10}\left(1-{\left(\frac{1}{10}\right)}^n\right)}{\frac{9}{10}}\\ S_n=1-\frac{1}{{10}^n}\\ S_n=\frac{{10}^n-1}{{10}^n}\end{array}$

Hence, the required answer is $S_n=\frac{{10}^n-1}{{10}^n}$ .

d.

To determine

Show that the formula is valid for all positive integral values of $n$ .

Answer to Problem 28E

The statement is valid for

  $k+1$ and for all integers.

Explanation of Solution

Given information:

  $\mathrm{0.99999....}=1$

Calculation:

Here, we have to prove:

  $S_n:\frac{9}{10}+\frac{9}{{10}^2}+\frac{9}{{10}^3}+\mathrm{.....}+\frac{9}{{10}^n}=\frac{{10}^n-1}{{10}^n}$ is valid for all integers $n$ .

Now, we will verify that the formula is valid for $n=1$

  $\begin{array}{l}L.H.S=\frac{9}{10}\\ R.H.S=\frac{10-1}{10}=\frac{9}{10}\end{array}$

Now, we will prove that it is true for $n=k+1$

Suppose $n=k$ :

  $\begin{array}{l}{S}_k=0.9+0.09+0.009+\mathrm{...}+\frac{9}{{10}^k}=\frac{{10}^k-1}{{10}^k}\\ \left(0.9+0.09+0.009+\mathrm{....}+\frac{9}{{10}^k}\right)+\frac{9}{{10}^{k+1}}=\left(\frac{{10}^k-1}{{10}^k}\right)+\frac{9}{{10}^{k+1}}\\ \therefore n=k\\ =\frac{10\left({10}^k-1\right)+9}{{10}^{k+1}}\\ =\frac{{10}^{k+1}-10+9}{{10}^{k+1}}\\ =\frac{{10}^{k+1}-1}{{10}^{k+1}}\end{array}$

Hence, the statement is valid for $k+1$ and for all integers.

e.

To determine

Show that $\mathrm{0.99999....}=1$ by using the formula.

Answer to Problem 28E

  $\mathrm{0.9999...}=1$

Explanation of Solution

Given information:

  $\mathrm{0.99999....}=1$

Calculation:

Here,

  $\begin{array}{l}\underset{n\to \infty }{{\displaystyle \lim }}\frac{{10}^n-1}{{10}^n}\\ \underset{n\to \infty }{{\displaystyle \lim }}\left(1-\frac{1}{{10}^n}\right)\\ \underset{n\to \infty }{{\displaystyle \lim }}\left(1\right)-\underset{n\to \infty }{{\displaystyle \lim }}\left(\frac{1}{{10}^n}\right)\\ 1-0=1\\ \mathrm{0.9999....}=1\end{array}$

Hence, $\mathrm{0.9999...}=1$ .