Question

Find sum to infinity of the series
1/8 - 1/9 +1/16 + 1/27 + 1/32 - 1/81 + .....................

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

For a Geometric Progression with

First term = a

Common Ratio = r ( < 1 )

Sum of the series upto infinite number of terms

$\displaystyle \: \sf{= \frac{a}{1 - r} }$

TO DETERMINE

The sum of series with infinite number of terms

$\displaystyle \: \sf{ \frac{1}{8} - \frac{1}{9} + \frac{1}{16} - \frac{1}{27} + \frac{1}{32} - \frac{1}{81} + ......... }$

CALCULATION

$\displaystyle \: \sf{ \frac{1}{8} - \frac{1}{9} + \frac{1}{16} - \frac{1}{27} + \frac{1}{32} - \frac{1}{81} + ......... }$

$\displaystyle \: \sf{ \bigg( \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + ......... \bigg) - \bigg( \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + ......... \bigg) \: }$

$= \sf{A - B \: \: \: (Say) }$

Where

$\displaystyle \: \sf{A = \bigg( \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + ......... \bigg) }$

$\displaystyle \: \sf{ B = \bigg( \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + ......... \bigg) \: }$

Now

$\displaystyle \: \sf{ \bigg( \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + ......... \bigg) } \: is \: a \: geometric \: series \:$

$\displaystyle \: \sf{First \: term = \frac{1}{8} \: and \: common \: ratio = \frac{1}{2} }$

So

$\displaystyle \: \sf{A= \frac{ \frac{1}{8} }{1 - \frac{1}{2} } = \frac{1}{8} \times 2 = \frac{1}{4} }$

Again

$\displaystyle \: \sf{ \bigg( \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + ......... \bigg) \: }is \: a \: geometric \: series \:$

$\displaystyle \: \sf{First \: term = \frac{1}{9} \: and \: common \: ratio = \frac{1}{3} }$

$\displaystyle \: \sf{B= \frac{ \frac{1}{9} }{1 - \frac{1}{3} } = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6} }$

So the required answer is

$\displaystyle \sf{ A - B = \frac{1}{4} - \frac{1}{6} = \frac{3 - 2}{12} = \frac{1}{12} \: }$