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

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Answer 1 · 2018-03-18T12:54:53

Convergent.

Sum to infinity #= 7/4#

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)#