How do you find the sum of the infinite geometric series #-3(3/4)^(n - 1)#?

1 Answer
Shwetank Mauria
Mar 5, 2016

Sum of the infinite geometric series #-3(3/4)^(n-1)# is #-12#.

Explanation:

Sum of a geometric series #{a,ar,ar^2,ar^3,.....,ar^(n-1)}# where #a# is first term and ratio is #r# (note #n^(th)# term is #ar^(n-1)#) is given by

#axx((r^n-1)/(r-1))# where #r>1# and #axx((1-r^n)/(1-r))# where #r<1#.

In the above case, #a=-3# and #n^(th)# term is #-3(3/4)^(n-1)#.

As #r<1# as #n->oo#, #r^n->0#.

Hence, sum of infinite series will be #(-3)xx1/(1-3/4)# or

#-3xx1/(1/4)# or #-3xx4=-12#

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