Un+1= aUn +b.
next term after Un is n+1/3^n+1
this makes:
Un+1= Un((n+1)/3n)
L= L((n+1)/3n)
L(1-((n+1)/3n))=0
for (n+1)/3n:
both n+1 and 3n have a limit of infinity, as is seen by examining them in recurrence relation format:
in n+1:
a is one, so will not influence terms but b, equal to one, a constant positive is being added to each term to make the next one, increasing the value of each successive term by an equal amount ,
in 3n:
, a is 1, so will not influence terms but b is 3, a constant postive is being added to each term to make the next one,increasing the value of each successive term by an equal amount. .
so the fraction, at limit, becomes infinity/infinity=0
this means :
L(1-0)=0
so L=0