Sum of infinite geometric series:

Any series in the form of $a,ar,a{r}^{2},...$ is called infinite geometric series and sum of infinite geometry series is,

${S}_{\mathrm{\infty}}=\frac{a}{1-r}$

Where a is the first term and r is a common ratio.

And Sum of geometric series:

${S}_{n}=\frac{a(1-{r}^{n})}{1-r}$

Given series,

$0.3+0.03+0.003+\cdots $

Since

$0.3=0.3$

$0.03=0.3(0.1)$

$0.003=0.3(0.1{)}^{2}$

So given series is a geometric series with first term $a=0.3$ and common ratio is 0.1.

So sum of infinite geometry series is,

${S}_{\mathrm{\infty}}=\frac{0.3}{1-0.1}$

${S}_{\mathrm{\infty}}=\frac{0.3}{0.9}$

${S}_{\mathrm{\infty}}=\frac{1}{3}$

${S}_{\mathrm{\infty}}=0.333...$

Hence

$0.3+0.03+0.003+...=0.333...$

Given series,

$0.9+0.09+0.009+0.0009+...$

Since

$0.9=0.9$

$0.09=0.9(0.1)$

$0.009=0.9(0.1{)}^{2}$

So given series is a geometric series with first term $a=0.9$ and common ratio is 0.1.

a) Evaluate the sum of the first one, two, three, and four terms of the series.

Since sum of geometric series:

${S}_{n}=\frac{a(1-{r}^{n})}{1-r}$

Since $a=0.9,r=0.1$

Sum of the first term,

${S}_{1}=\frac{0.9(1-(0.1{)}^{1})}{1-(0.1)}=\frac{0.9(0.9)}{0.9}=0.9$

Sum of the first two terms,

${S}_{2}=\frac{0.9(1-(0.1{)}^{2})}{1-(0.1)}=\frac{0.9(0.99)}{0.9}=0.99$

Sum of the first three terms,

${S}_{3}=\frac{0.9(1-(0.1{)}^{3})}{1-(0.1)}=\frac{0.9(0.999)}{0.9}=0.999$

Sum of the first four terms,

${S}_{4}=\frac{0.9(1-(0.1{)}^{4})}{1-(0.1)}=\frac{0.9(0.9999)}{0.9}=0.9999$

Hence

Sum of the first term $=0.9$

Sum of the first two terms $=0.99$

Sum of the first three terms $=0.999$

Sum of the first four terms $=0.9999$

b) So sum of infinite geometry series is,

${S}_{\mathrm{\infty}}=\frac{0.9}{1-0.1}$

${S}_{\mathrm{\infty}}=\frac{0.9}{0.9}$

${S}_{\mathrm{\infty}}=1$

Hence

$0.9+0.09+0.009+0.0009+...=1$