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I have a calculus question?
Show that the sequence defined by a1 = 2, a(n +1) = (1/[3-an]) satisfies 0 < an <= 2 and is decreasing. Deduce that the sequence is convergent and find it’s limit.
1 Answer
- ?Lv 61 decade agoFavorite Answer
a1 = 2
a(n+1) = 1/(3-an) , where 0<an<=2
You can show the sequence is convergent and is decreasing by find a few numbers.
a2 = a(1+1) = 1/(3-a1) = 1/(3-2) = 1/1 = 1
a3 = a(2+1) = 1/(3-a2) = 1/(3-1) = 1/2
a4 = a(3+1) = 1/(3-a3) = 1/(3-0.5) = 2/5
a5 = a(4+1) = 1/(3-a4) = 1/(3-0.4) = 5/13
And yes you can see that it is decreasing and getting closer to 0.
lim(n->infi) [1/(3-an)] = 1/infi = 0