Infinite geometric series? #sum_(n=1) ^oo (3^(2n))/4^n#

Does this series diverge? How would you rearrange it to solve?

#sum_(n=1) ^oo (3^(2n))/4^n#

1 Answer
Mar 31, 2018

Diverges.

Explanation:

So, the first objective is to get this in the form

#sum_(n=0)^ooa(r)^n# where #a# is just a constant (and the first term of the series) and #r# is the common ratio raised to the #nth# power.

So, let's take #a_n=3^(2n)/4^n# and rewrite as #sum_(n=1)^oo(3^2)^n/4^n#, as #x^(an)=(x^a)^n#.

This simplifies to #a_n=9^n/4^n=(9/4)^n#

We then have

#sum_(n=1)^oo(9/4)^n#. We can shift the index (starting value) to #0,# which requires adding #1# to all #n# in #a_n=(9/4)^n#:

#sum_(n=0)^oo(9/4)^(n+1)=sum_(n=0)^oo(9/4)(9/4)^n#

So, #a=9/4, r=9/4#.

Now, for a geometric series #sum_(n=0)^ooa(r)^n,# the series converges only if #|r|<1#. Here, #|r|=|9/4|>1#, so the series diverges.