Quiz 6 Solutions, Math 112, Section 2 (Vinroot)
For each of the following, determine if the sequence converges or diverges, and if the sequence
converges, find its limit. Give a brief explanation for each (a short sentence will do).
(a): an= (−2)n.
Solution: Since angets larger in absolute value as ngets large, the sequence diverges. (Also, the
sign of analternates, but the fact that the absolute value of angets arbitrarily large is enough to
conclude that the sequence diverges.)
(b): bn= 3 −(1/2)n.
Solution: We have
lim
n→∞
bn= lim
n→∞
(3 −(1/2)n) = lim
n→∞
3−lim
n→∞
(1/2)n= 3 −0 = 3,
since the sequence rnconverges to 0 when |r|<1. Thus, the sequence converges to 3.
(c): cn= (5n3+ 2)/3n.
Solution: Consider the function f(x) = (5x3+ 2)/3x, which satisfies f(n) = cnfor any positive
integer n. Since both 5x3+ 2 and 3xgo to +∞as x→ ∞, we may apply L’Hospital’s rule to find
the limit of f(x) as x→ ∞. When we calculate this limit, we must actually apply L’Hospital’s rule
3 times (since the numerator is a polyonmial of degree 3):
lim
x→∞
((5x3+ 2)/3x) = lim
x→∞
(15x2/((ln 3)3x)) = lim
x→∞
(30x/((ln 3)23x)) = lim
x→∞
(30/((ln 3)33x)) = 0.
Since f(x) converges to 0 as x→ ∞, then cnalso converges to 0.
(d): dn= 1 + (cos(n)/n2).
Solution: For any n, cos(n) is no larger than 1 and no smaller than −1. So, for any n≥1,
−1/n2≤cos(n)/n2≤1/n2.
We also know that
lim
n→∞
(−1/n2) = 0 and lim
n→∞
(1/n2) = 0.
Therefore, by the Squeeze Theorem for sequences,
lim
n→∞
(cos(n)/n2)=0.
For the sequence dn, then, we have
lim
n→∞
dn= lim
n→∞
(1 + (cos(n)/n2)) = lim
n→∞
1 + lim
n→∞
(cos(n)/n2) = 1 + 0 = 1.
Thus, the sequence {dn}converges to 1.