Question #41dc8

1 Answer
Nov 14, 2017

See below.

Explanation:

This is a geometric series. The sum to infinity of a geometric series can be expressed as:

#a((1-r^n)/(1-r))#

Where #a# is the first term, #r# is the common ratio and n is the nth term.

The limit to infinity of:

#lim_(n->oo)(a-r^n)=a# if and only if #-1 < r < 1#

( since #r^n-> 0#, this is convergence)

If #color(white)(88)1 < r < -1#

Then:

#r > 1color(white)(88)#

#lim_(n->oo)(a-r^n)=-oo# ( this is divergence )

( for #r < -1 # the limit is undefined )

So from example:

Common difference is:

#(1/3)/(-1/sqrt(3))=(-1/(3sqrt(3)))/(1/3)=(-sqrt(3))/3#

#-1 < (-sqrt(3))/3 < 1# ( so this is a convergent series )

Sum to infinity:

#-1/sqrt(3)(1/(1-(-sqrt(3))/3))=((-1)/sqrt(3))/(1+(sqrt(3))/3)=-sqrt(3)/(3+sqrt(3))#