Find the sum of the series #1/2-1/3+1/4-1/9+1/8-1/27+........# to infinity?

1 Answer
Jan 16, 2017

#1/2-1/3+1/4-1/9+1/8-1/27+........=1/2#

Explanation:

#1/2-1/3+1/4-1/9+1/8-1/27+........# can be rewritten as

#(1/2+1/4+1/8+........)-(1/3+1/9+1/27+........)#

Hence, the given infinitive is one infinitive geometric series subtracted from another infinitive geometric series.

Observe that in the infinite series #1/2+1/4+1/8+........#, while first term #1/2# and common ratio is #1/2#

and in the infinite series #1/3+1/9+1/27+........#, first term #1/3# and common ratio is #1/3#

As for an infinite series whose first term is #a# and common ratio #r# is such that #|r|<1#, the sum converges to #a/(1-r)#.

Hence #1/2+1/4+1/8+........=(1/2)/(1-1/2)=(1/2)/(1/2)=1#, and

#1/3+1/9+1/27+........=(1/3)/(1-1/3)=(1/3)/(2/3)=1/2#

Hence #1/2-1/3+1/4-1/9+1/8-1/27+........=1-1/2=1/2#