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Solutions to quiz 8 of the calculus ii (math 106a) course offered at the university of california, berkeley, in the fall of 2005. The solutions cover two problems: finding the limit of a sequence using l'hopital's rule and determining the convergence of a geometric series.
Typology: Exercises
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QUIZ 8
Show ALL your work CAREFULLY.
(a) Find the limit of the following sequence or explain why it does not exist. an = n(1 − cos(1/n)) Note that as n → ∞, cos(1/n) → 1 so that an approaches to an indeterminate form of ∞ · 0. We need to re-write an and then use L’Hˆopital’s rule. Write an = (1−cos(1 (1/n)/n )). Then
lim n→∞
an = lim n→∞
(1 − cos(1/n)) (1/n)
= (^) nlim→∞
sin(1/n) · (− 1 /n^2 ) −(1/n^2 )
by L’Hˆopital’s rule
= (^) nlim→∞ sin(1/n) = 0.
Thus the sequence {an} converges to 0. (b) Determine whether the following infinite series converges or diverges. If it converges, evaluate the infinite series.
∑^ ∞
n=
( (^) π 4
)n
This is a geometric series where the leading term a = 3
(π 4
with common ratio r = π 4. Since π 4 < 1 , the geometric series converges and thus (^) ∞ ∑
n=
( (^) π
4
a 1 − r
(π 4
1 − π 4
=
3 π^3 16(4 − π)
Date: November 14, 2005. 1