Question

Find the sum of the following series to infinity :

(i) $1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}+....\infty$

(ii) $8+4\sqrt{2}+4+....\infty$

(iii) $\frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+.....\infty$.

(iv) 10 - 9 + 8.1 - 7.29 + ...... $\infty$

(v) $\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+.....\infty$

Answer 1

(i) $S_{\infty }=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....$

$\Rightarrow a=1,r=-\frac{1}{3}$

$S_{\infty }=\frac{a}{1-r}=\frac{1}{1+\frac{1}{3}}$

$S_{\infty }=\frac{3}{4}$.

(ii) $S_{\infty }=8+4\sqrt{2}+4+....$

$\Rightarrow a=8,r=-\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$

$S_{\infty }=\frac{a}{1-r}=\frac{8}{1-\frac{1}{\sqrt{2}}}$

$=\frac{8\sqrt{2}}{\sqrt{2}-1}\times \frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}$

$=\frac{8(2+\sqrt{2})}{2-1}$

$S_{\infty }=8(2+\sqrt{2})$

(iii) $S_{\infty }=\frac{2}{5}+\frac{2}{5^2}+\frac{2}{5^3}+\frac{2}{5^4}+........$

$=(\frac{2}{5}+\frac{3}{5^3}+...)+(\frac{3}{5^2}+\frac{3}{5^4}+....)S_{\infty }=S_{\infty }^{\prime }+S{"}_{\infty }For{S}_{\infty }^{\prime }=\frac{a}{1-r}=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{2}{5}\times \frac{25}{24}=\frac{10}{24}{S}_{\infty }^{\prime }=\frac{5}{12}S{"}_{\infty }=\frac{\frac{3}{25}}{1-\frac{1}{25}}=\frac{3}{25}\times \frac{25}{24}=\frac{3}{24}{S}_{\infty }=S_{\infty }^{\prime }+S{"}_{\infty }=\frac{10}{24}+\frac{3}{24}+\frac{13}{24}$

(iv) This infinite G.P. has first term a = 10 and common ratio = - $\frac{9}{10}=-0.9$

Thus the sum of the infinite G.P. will be :

$10-9+8.9-7.29+....\infty \frac{a}{1-r}[Since|r|<1]$

$=\frac{10}{1-(-0.9)}$

$=\frac{10}{1.9}=\frac{100}{19}$

(v) $\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+....\infty$

The G.P. can be written as follows :

$\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+.....\infty$

$=(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+....\infty )+(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+....\infty )=\frac{\frac{1}{3}}{1-\frac{1}{3^2}}+\frac{\frac{1}{5^2}}{1-\frac{1}{5^2}}=\frac{3}{8}+\frac{1}{24}=\frac{10}{24}=\frac{5}{12}$