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))#
