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Determine whether the sequence converges or diverges
Determine whether the sequence converges or diverges. If converges find the limit.
A) an = (n + 2)!/n!
B) {n^2e^-n}
C) an = ln(2n^2 + 1)-ln(n^2 + 1)
D) an = cos^2n/2^n
Expert Solution
A) Diverge.
Because an = (n+2)!/n! = (n+2)(n+1) -> oo as n->oo
B) Converge, limit is 0.
We consider f(x)=x^2*e^(-x)=x^2/e^x. Let x->oo, then we have
lim f(x) = lim (x^2)'/(e^x)'=lim 2x/e^x
= lim (2x)'/(e^x)' = lim 2/e^x = 0
Thus an=n^2*e^(-n) -> 0 as n->oo.
C) Converge, limit is ln2.
exp(an) = exp(ln(2n^2+1)-ln(n^2+1))
= (2n^2+1)/(n^2+1) -> 2 as n->oo
So the limit of an is ln2
D) Converge, limit is 0
Because |an|=|cos^2n/2^n|<=1/2^n->0 as n->oo. Thus the limit
of an is 0.
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