66 views

Consider the series $\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{8}+\frac{1}{9}-\frac{1}{16}+\frac{1}{32}+\frac{1}{27}-\frac{1}{64}+\ldots.$ The sum of the infinite series above is$:$

1. $\infty$
2. $\frac{5}{6}$
3. $\frac{1}{2}$
4. $0$

retagged | 66 views

We can observe that there is sum of two series.

$\color{red}{\left( \;\frac{(1)}{2} \;+ \;\frac{(-1)}{2^2} \;+ \;\frac{(1)}{2^3} \;+ \;\frac{(-1)}{2^4} \;+ \;\frac{(1)}{2^5} \;+ \;\frac{(-1)}{2^6} \;+ \;\frac{(1)}{2^7} \;+\ldots\right)}$$+ \color{magenta}{\left( \;\frac{1}{3^1} \;+\;\frac{1}{3^2} \;+\;\frac{1}{3^3} \;+\;\frac{1}{3^4} \;+\;\frac{1}{3^5} \;+\ldots\right) }$

Both parts are in G.P.
First series with $a = \frac{1}{2}$ and $r = \frac{(-1)}{2} \implies$ infinite sum $= \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{(-1)}{2}} =\frac{1}{3}$

Second series with $a = \frac{1}{3}$ and $r = \frac{(1)}{3} \implies$ infinite sum $= \frac{a}{1-r} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}$

Final Sum $=\frac{1}{3} + \frac{1}{2} = \frac{5}{6}.$
by Veteran (62k points)
edited by
+1 vote

$\frac{1}{2}$ + $\frac{1}{3}$- $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{9}$ - $\frac{1}{16}$ + $\frac{1}{32}$ +$\frac{1}{27}$ - $\frac{1}{64}$....

$\Rightarrow$ $\frac{{\color{Red} 1}}{{\color{Red} 2}}$ + $\frac{{\color{Green} 1}}{{\color{Green} 3}}$ - $\frac{1}{4}$ + $\frac{{\color{Red} 1}}{{\color{Red} 8}}$ + $\frac{{\color{Green} 1}}{{\color{Green} 9}}$ - $\frac{1}{16}$+ $\frac{{\color{Red} 1}}{{\color{Red} 3 \color{Red}2}}$ + $\frac{{\color{Green} 1}}{{\color{Green} 2\color{Green} 7}}$ - $\frac{1}{64}$.....

$\Rightarrow$ $\frac{{\color{Red} 1}}{{\color{Red} 2}}$ + $\frac{{\color{Red} 1}}{{\color{Red} 8}}$ + $\frac{{\color{Red} 1}}{{\color{Red} 3 \color{Red}2}}$ +... +$\frac{{\color{Green} 1}}{{\color{Green} 3}}$+ $\frac{{\color{Green} 1}}{{\color{Green} 9}}$+ $\frac{{\color{Green} 1}}{{\color{Green} 2\color{Green} 7}}$+...- $\frac{1}{4}$- $\frac{1}{16}$- $\frac{1}{64}$-...

$\Rightarrow$ $\left ( \frac{{\color{Red} 1}}{{\color{Red} {2^{1}}}}+ \frac{{\color{Red} 1}}{{\color{Red}{ 2^{3}}}} + \frac{{\color{Red} 1}}{\color{Red}{2^{5}}}+...\right )$ + $\left ( \frac{{\color{Green} 1}}{{\color{Green} 3}}+ \frac{{\color{Green} 1}}{{\color{Green} { 3^{2}}}}+ \frac{{\color{Green} 1}}{{\color{Green}{ 3^{3}} }} +... \right )$-$\left (\frac{1}{2^{2}}+ \frac{1}{2^{4}}+\frac{1}{2^{6}}+... \right )$

Hence the answer is the sum of 3 infinite G.P. series.

Sum of infinite G.P. Series = $\frac{a}{1-r}$

$\left ( \frac{{\color{Red} 1}}{{\color{Red} {2^{1}}}}+ \frac{{\color{Red} 1}}{{\color{Red}{ 2^{3}}}} + \frac{{\color{Red} 1}}{\color{Red}{2^{5}}}+...\right )$ = $\frac{{\color{Red} 1}}{{\color{Red} 2}}\left ( {\color{Red} 1} + \frac{{\color{Red} 1}}{{\color{Red} {2^{2}}}} + \frac{{\color{Red} 1}}{{\color{Red} {2^{4}}}} + ...\right )$ = $\frac{{\color{Red} 1}}{{\color{Red} 2}}\left ( \frac{\color{Red}1}{{\color{Red} {1-\frac{1}{4}}}} \right )$ =$\frac{{\color{Red} 1}}{{\color{Red} 2}} * \frac{{\color{Red} 4}}{{\color{Red} 3}}$ = $\frac{{\color{Red} 2}}{{\color{Red} 3}}$

$\left ( \frac{{\color{Green} 1}}{{\color{Green} 3}}+ \frac{{\color{Green} 1}}{{\color{Green} { 3^{2}}}}+ \frac{{\color{Green} 1}}{{\color{Green}{ 3^{3}} }}+... \right )$ = $\frac{{\color{Green} 1}}{{\color{Green} 3}} \left ( {\color{Green} 1}+ \frac{{\color{Green} 1}}{{\color{Green} 3}} + \frac{{\color{Green} 1}}{{\color{Green} {3^{2}}}} +...\right )$ = $\frac{{\color{Green} 1}}{{\color{Green} 3}} \left ( \frac{{\color{Green} 1}}{{\color{Green} {1 - \frac{1}{3} }}} \right )$ = $\frac{{\color{Green} 1}}{{\color{Green} 3}} * \frac{{\color{Green} 3}}{{\color{Green} 2}}$ = $\frac{{\color{Green} 1}}{{\color{Green} 2}}$

$\left (\frac{1}{2^{2}}+ \frac{1}{2^{4}}+\frac{1}{2^{6}}+... \right )$ = $\frac{1}{4} \left (1+ \frac{1}{2^{2}}+\frac{1}{2^{4}}+... \right )$ = $\frac{1}{4}\left ( \frac{1}{1-\frac{1}{4}} \right )$ = $\frac{1}{4} * \frac{4}{3}$ = $\frac{1}{3}$

$\frac{{\color{Red} 2}}{{\color{Red} 3}} + \frac{{\color{Green} 1}}{{\color{Green} 2}} - \frac{1}{3}$ = $\frac{{\color{Red} 4 }+ {\color{Green} 3} -2}{6}$ = $\frac{5}{6}$

$\therefore$ Option B   $\frac{5}{6}$ is the right answer.

by Boss (17.5k points)
edited by
$\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{8}+\frac{1}{9}-\frac{1}{16}+\frac{1}{32}+...$

= $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{32}+\frac{1}{3}+\frac{1}{9}...$

=$\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...+\frac{1}{3}+\frac{1}{9}+...$

$\frac{\frac{1}{4}}{1-\frac{1}{4}}$+$\frac{\frac{1}{3}}{1-\frac{1}{3}}$

=$\frac{1}{4} \times \frac{4}{3} + \frac{3}{2}\times\frac{1}{3}$

=$\frac{1}{3} + \frac{1}{2}$= $\frac{2}{6}+\frac{3}{6}$=$\frac{5}{6}$
by Active (2.1k points)