How do you determine if #a_n=1+3/7+9/49+...+(3/7)^n+...# converge and find the sums when they exist?

Question

Paisley Johnson · Accepted Answer

Convergent.

Sum to infinity #= 7/4# This is a geometric series: The #bb(nth)# term in a geometric series is given by: #ar^(n-1)# Where: #bba# is the first term, #bbr# is the common ratio and #bbn# is the #bb(nth)# term. If: #a , b , c # are in geometric sequence, then: #b/a=c/b# This is known as the common ratio. So we have: #(3/7)/1=(9/49)/(3/7)=3/7# Common ratio #bb(3/7)# First term is #bb1# The sum of a geometric series is given by: #a((1-r^(n))/(1-r))# From this we can see that if: #|r|<1# then: #lim_(n->oo)a((1-r^(n))/(1-r))=a/(1-r)# This is a finite value, and the sum is said to converge. If: #|r|>1# then: The sum increases without bound, and is said to diverge. We have #r=3/7# #|3/7|<1# so the series converges. #sum_(n=1)^(oo)(3/7)^(n-1)=1/(1-3/7)=color(blue)(7/4)#

Luke Alvarez · Answer

undefined This is a geometric series with first term $ a_0 = 1 $ and common ratio $ r = \frac{3}{7} $. The series converges if $ |r| < 1 $. In this case, $ |3/7| < 1 $ so the series converges.

The sum of a convergent geometric series is given by the formula $ S = \frac{a}{1 - r} $, where $ a $ is the first term and $ r $ is the common ratio.

Substituting the values, we get $ S = \frac{1}{1 - \frac{3}{7}} = \frac{1}{\frac{4}{7}} = \frac{7}{4} $. Thus, the sum of the series is $ \frac{7}{4} $.