Problem 648

Find the sum of the following infinite geometric series, if it exists.

 \frac{1}{2}+\left ( \frac{-1}{4} \right )+\frac{1}{8}+\left ( \frac{-1}{16}  \right )+...

 

Solution:-

 

S = \frac{a}{1-r} , where r = common ratio (r ≤ 1) and a = first term.

a = \frac{1}{2} , r = -1/8

S =  \frac{\frac{1}{2}}{\left ( 1-\left ( \frac{-1}{8} \right ) \right )}

S = \frac{4}{9}

So the sum of the infinite geometric series 4/9
 

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