How do you use the ratio test to test the convergence of the series #∑ (3/4)^n# from n=1 to infinity?

Question

3793 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-12T14:21:25

The series is convergent and:

#sum_(n=1)^(+oo) (3/4)^n = 3#

The ratio to test is:

#r= a_(n+1)/a_n= frac ((3/4)^(n+1)) ((3/4)^n) =3/4#

As #r<1# the series is convergent.

We can note that this is a particular case of the geometric series:

#sum_(n=0)^(+oo) x^n = 1/(1-x)# for #|x|<1#.

So that we can also calculate the sum:

#sum_(n=1)^(+oo) (3/4)^n = -1 + sum_(n=0)^(+oo) (3/4)^n = -1+1/(1-3/4) = -1 + 1/(1/4)=-1+4=3 #