Differentiation (Part-38) | S N De

 $\,1.\,$ Find the sum of each f the following infinite geometric series, if it exists :

$\,(i)\,1+\frac 13+\frac 19+\frac{1}{27}+\cdots \infty$

Sol. Let $\,S_1=1+\frac 13+\frac 19+\frac{1}{27}+\cdots \infty$ be an infinite series whose first term $=1,\,$ and common ratio $=\frac 13.$

So, $\,S_1=1+\frac 13+\left(\frac 13\right)^2+\left(\frac 13\right)^3+\cdots \infty\\~~~~~=\frac{1}{1-\frac 13}\\~~~~~=\frac{1}{2/3}\\~~~~~=\frac 32\,\,\text{(ans.)}$

$\,(ii)\,\frac 13-\frac 29+\frac{4}{27}-\frac{8}{81}+\cdots \infty $

Sol. Let $\,S_1=\frac 13-\frac 29+\frac{4}{27}-\frac{8}{81}+\cdots \infty$ be an infinite series whose first term $=\frac 13,\,$ and common ratio $=\frac{-2/9}{1/3}=-\frac 23.$

So, $\,S_1=\frac 13-\frac 29+\frac{4}{27}-\frac{8}{81}+\cdots \infty\\~~~~~=\frac 13 \times \frac{1}{1-\left(-\frac 23\right)}\\~~~~~=\frac{1}{3}\times \frac{1}{1+(2/3)}\\~~~~~=\frac {1}{3}\times \frac{1}{5/3}\\~~~~~=\frac 13 \times \frac 35\\~~~~~=\frac 15\,\,\text{(ans.)}$

$\,(iii)\,\frac 12+\frac 34+\frac{9}{8}+\frac{27}{16}+\cdots \infty$

Sol. Let $\,S_1=\frac 12+\frac 34+\frac{9}{8}+\frac{27}{16}+\cdots \infty $ be an infinite series whose first term $=\frac 12,\,$ and common ratio $(r)=\frac{3/4}{1/2}=\frac 32(>1).$

So,as we know that the infinite geometric series has no sum when $\,r>1\,\,$ and hence the given series $\,S_1\,\,$ has no sum.

$\,(iv)\,1+1+1+1+\cdots \infty$

Sol.  Let $\,\,S_1=1+1+1+1+\cdots \infty .\,$ 

The given infinite geometric series has the first term $=1,\,$ and the common ratio $(r)=1.$

Since $\,r \not\in (-1,1),~$ the given infinite geometric series has no sum. 

$\,(v)\,\frac 25+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{2}{5^5}+\frac{3}{5^6}+\cdots \infty$

Sol.  Let $\,\,S=\frac 25+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{2}{5^5}+\frac{3}{5^6}+\cdots \infty \\~~~~=\left(\frac 25+\frac{2}{5^3}+\frac{2}{5^5}+\cdots\infty\right)\\~~~~~+\left(\frac{3}{5^2}+\frac{3}{5^4}+\frac{3}{5^6}+\cdots \infty\right)\\~~~~=S_1+S_2,\,\,\text{(say)}$ 

Now, for the infinite geometric series $\,S_1,\,$ the first term $=\frac 25,\,$ common ratio $(r)=\frac{2/5^3}{2/5}=\frac{1}{5^2}.$

So, $\,\,S_1=\frac 25+\frac{2}{5^3}+\frac{2}{5^5}+\cdots\infty\\~~~~~=\frac 25 \cdot\frac{1}{1-\frac{1}{5^2}}\\~~~~=\frac 25 \cdot \frac{1}{\frac{24}{25}}\\~~~~=\frac 25 \cdot \frac{25}{24}\\~~~~=\frac{5}{12}.$

Again, for the infinite geometric series $\,S_2,\,$ the first term $=\frac{3}{5^2},\,$ common ratio $(r)=\frac{3/5^4}{3/5^2}=\frac{1}{5^2}.$

So, $\,\,S_2=\frac{3}{5^2}+\frac{3}{5^4}+\frac{3}{5^6}+\cdots \infty\\~~~~~=\frac{3}{5^2} \cdot\frac{1}{1-\frac{1}{5^2}}\\~~~~=\frac{3}{5^2} \cdot \frac{1}{\frac{24}{25}}\\~~~~=\frac {3}{25} \cdot \frac{25}{24}\\~~~~=\frac{1}{8}.$

Hence, $\,S=S_1+S_2\\~~~=\frac{5}{12}+\frac 18\\~~~=\frac{10+3}{24}\\~~~=\frac{13}{24}.$

$\,(vi)\,\, 0.9+0.81+0.729+\cdots\infty$

Sol. Let the given infinite geometric series be denoted by $\,S\,$, where the first term $=0.9,\,$ common ratio $(r)=\frac{0.81}{0.9}=0.9(<1)$.

Now, sum of the given series exist since $\,-1<r<1.$

So, $\,S=0.9+0.81+0.729+\cdots\infty \\~~~=0.9+(0.9)^2+(0.9)^3+\cdots \infty\\~~~=0.9 \times \frac{1}{1-0.9}\\~~~=\frac{0.9}{0.1}\\~~~=9\,\,\,\text{(ans.)}$

$\,(vii)\,\frac{1}{16}+\frac{1}{12}+\frac{1}{9}+\frac{4}{27}+\cdots \infty$

Sol. Let $\,S_1=\frac{1}{16}+\frac{1}{12}+\frac{1}{9}+\frac{4}{27}+\cdots \infty $ be an infinite series whose first term $=\frac{1}{16},\,$ and common ratio $(r)=\frac{1/12}{1/16}=\frac 43(>1).$

So,as we know that the infinite geometric series has no sum when $\,r>1\,\,$ and hence the given series $\,S_1\,\,$ has no sum.

$\,(viii)\,~~\frac 47-\frac{5}{7^2}+\frac{4}{7^3}-\frac{5}{7^4}+\frac{4}{7^5}-\frac{5}{7^6}+\cdots \infty$

Sol. Let $\,\,S=\frac 47-\frac{5}{7^2}+\frac{4}{7^3}-\frac{5}{7^4}+\frac{4}{7^5}-\frac{5}{7^6}+\cdots \infty\\=(\frac{4}{7}+\frac{4}{7^3}+\frac{4}{7^5}+\cdots \infty)-(\frac{5}{7^2}+\frac{5}{7^4}+\frac{5}{7^6}+\cdots \infty)\\=S_1-S_2,\,\,\text{(say).}$

Now, $\,S_1~=\frac{4}{7}+\frac{4}{7^3}+\frac{4}{7^5}+\cdots \infty\\~~~~~~=\frac 47 \times \frac{1}{1-\frac{1}{7^2}}\\ ~~~~~~~~~~[\text{first term}(a)=\frac 47, \text{common ratio}(r)=\frac{4/7^3}{4/7}=\frac{1}{7^2}]\\~~~~~~=\frac 47 \times \frac{1}{48/49}\\~~~~~~=\frac 47 \times \frac{49}{48}\\~~~~~~=\frac{7}{12}$

Again, $\,S_2=\frac{5}{7^2}+\frac{5}{7^4}+\frac{5}{7^6}+\cdots \infty\\~~~~~~=\frac{5}{7^2} \times \frac{1}{1-\frac{1}{7^2}}\\ ~~~~~~~~~~[\text{first term}(a)=\frac{5}{7^2}, \text{common ratio}(4)=\frac{5/7^4}{5/7^2}=\frac{1}{7^2}]\\~~~~~~=\frac{5}{7^2} \times \frac{1}{48/49}\\~~~~~~=\frac {5}{49}\times \frac{49}{48}\\~~~~~~=\frac{5}{48}$

Hence, $\,\,~S~=S_1-S_2\\~~~~~~=\frac{7}{12}-\frac{5}{48}\\~~~~~~=\frac{28-5}{48}\\~~~~~~=\frac{23}{48}\,\,\text{(ans.)}$

$\,2.\,$ Express each of the following recurring decimals  in the form of an infinite geometric series and obtain their values in the form of rational numbers :

$\,(a)\,0.\overline{3}$

Sol. Let $\,S=0.\overline{3}\\~~~=0.33333\cdots\\~~~=0.3+0.03+0.003+\cdots\\~~~=\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\cdots \rightarrow(1)$

Clearly, $\,(1)\,$ represents an infinite geometric series with the first term $(a)=\frac{3}{10},\,$ common ratio $(r)=\frac{3/10^2}{3/10}=\frac{1}{10}.$

Hence, $\,S=\frac{3}{10} \times \frac{1}{1-\frac{1}{10}}\\~~~=\frac{3}{10}\times \frac{1}{9/10}\\~~~=\frac{3}{10}\times \frac{10}{9}\\~~~=\frac 39\\~~~=\frac 13\,\,\text{(ans.)}$

$\,(b)\,0.3\overline{15}\\=0.3+0.015+0.00015+0.0000015+\cdots\\=0.3+\frac{15}{10^3}+\frac{15}{10^5}+\frac{15}{10^7}+\cdots\\=0.3+\frac{15}{10^3}\times \frac{1}{1-\frac{1}{10^2}}\\=\frac{3}{10}+\frac{15}{10^3}\times \frac{1}{99/10^2}\\=\frac{3}{10}+\frac{15}{10^3}\times \frac{10^2}{99}\\=\frac{3}{10}+\frac{5}{10 \times 33}\\=\frac{3}{10}+\frac{1}{66}\\=\frac{99+5}{330}\\=\frac{104}{330}\\=\frac{52}{165}\,\,\text{(ans.)}$

$\,(c)\,3.17\overline{35}\\=3.17353535\cdots\\=3.17+0.0035+0.000035+\cdots\\=3.17+\frac{35}{10^4}+\frac{35}{10^6}+\cdots\\=3.17+\frac{35}{10^4} \times \frac{1}{1-\frac{1}{10^2}}\\~~~~~[\because ~\text{common ratio}(r)=\frac{35/10^6}{35/10^4}=\frac{1}{10^2}]\\=3.17+\frac{35}{10^4}\times \frac{10^2}{99}\\=\frac{317}{100}+\frac{35}{9900}\\=\frac{317 \times 99+35}{9900}\\=\frac{31418}{9900}\\=\frac{15709}{4950}\,\,\text{(ans.)}$

$\,(d)\,1.\overline{41}+2.2\overline{12}\\=1.414141\cdots+2.2121212\cdots\\=(1+0.41+0.0041+0.000041+\cdots)\\~~~~+(2.2+0.012+0.00012+\cdots)\\=\left(1+\frac{41}{10^2}+\frac{41}{10^4}+\cdots \right)\\~~~~~~~+\left(2.2+\frac{12}{10^3}+\frac{12}{10^5}+\cdots\right)\\=(1+S_1)+(2.2+S_2)\,\,\text{(say)}$

Clearly, $\,S_1\,$ denotes an infinite geometric series with first term $=\frac{41}{10^2},\,\,$ and common ratio $(r)=\frac{41/10^4}{41/10^2}=\frac{1}{10^2}.$

So, $\,S_1=\frac{41}{10^2}\times \frac{1}{1-\frac{1}{10^2}}\\~~~~~=\frac{41}{10^2}\times \frac{10^2}{99}\\~~~~~=\frac{41}{99}$

Again, $\,S_2\,$ denotes an infinite geometric series with first term $=\frac{12}{10^3},\,\,$ and common ratio $(r)=\frac{12/10^5}{12/10^3}=\frac{1}{10^2}.$

So, $\,S_2=\frac{12}{10^3}\times \frac{1}{1-\frac{1}{10^2}}\\~~~~~=\frac{12}{10^3}\times \frac{10^2}{99}\\~~~~~=\frac{12}{990}$

Hence, the desired result

$=1+S_1+2.2+S_2\\=1+\frac{41}{99}+\frac{22}{10}+\frac{12}{990}\\=\left(1+\frac{22}{10}\right)+\frac{41}{99}+\frac{12}{990}\\=\frac{32}{10}+\frac{41}{99}+\frac{12}{990}\\=\frac{32\times 99+41 \times 10+12}{990}\\=\frac{3590}{990}\\=\frac{359}{99}\,\,\text{(ans.)}$