Calculus II Quiz 8 Solutions - Fall 2005 (MATH 106A), Exercises of Calculus

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.

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MATH 106A - CALCULUS II
FALL 2005
QUIZ 8
NAME:
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)1so that anapproaches to an
indeterminate form of ∞·0. We need to re-write anand then use
L’Hˆopital’s rule. Write an=(1cos(1/n))
(1/n). Then
lim
n→∞
an= lim
n→∞
(1 cos(1/n))
(1/n)
= lim
n→∞
sin(1/n)·(1/n2)
(1/n2)by L’Hˆopital’s rule
= lim
n→∞
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.
X
n=3
3π
4n
This is a geometric series where the leading term a=3π
43with
common ratio r=π
4. Since π
4<1, the geometric series converges
and thus
X
n=3
3π
4n=a
1r
=3π
43
1π
4
=3π3
16(4 π).
Date: November 14, 2005.
1

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MATH 106A - CALCULUS II

FALL 2005

QUIZ 8

NAME:

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

)n

a 1 − r

(π 4

1 − π 4

=

3 π^3 16(4 − π)

Date: November 14, 2005. 1