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MATH 222
(31)

Christa Scholtz
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Midterm

School

McGill University
Department

Mathematics & Statistics (Sci)

Course Code

MATH 222

Professor

Christa Scholtz

Description

McGill University
Department of Mathematics and Statistics
Calculus 3 (MATH 222)
Fall 2008
Instructors: Prof. K. GowriSankaran, N. Sonnerat
Midterm Exam
Solutions October 22nd 2008, 6pm - 8pm
Instructions: This is a 2 hour, closed book/notes examination. No calculators are permitted.
To get full marks, you must answer all ﬁve questions. Justify all your answers, unless the
question instructs you not to.
(−1) en
1. (a) Compute lim n→∞ { n! }.
Solution: The easiest way would have been to say that you saw in class that
xn
lim n! = 0 for all values of x, so in particular this is true for x = −e. Alternatively,
you can use the ratio test to show that the series
X (−1) e n
n!
converges absolutely. Then the test for divergence implies that the sequence must
have limit 0. As a third option, you could have shown that
(−1) e n en e2e
0 ≤ | | = ≤ ,
n! n! 2 n
so lim|an| = 0 by the squeeze theorem (see assignment solutions for more details).
But you have another theorem that tells you that if |a | tends to 0, then a must
n n
also tend to 0.
(b) Determine whether the series
X∞ 2 k!
k=0 (k + 2)!
converges or diverges.
Solution: Either use l’Hopital’s Rule to show that
2 k! 2k
lim = lim 6= 0,
(k + 2)! (k + 2)(k + 1)
which implies that the series is divergent by the test for divergence. Note that
x x x−1
d/dx(2 ) = ln2 · 2 , NOT x2 .
1 Alternatively, apply the ratio test to ﬁnd that
|an+1|
lim = 2 > 1,
|an|
so the series must diverge.
(c) Determine whether the series
√
X j
(−1) j+1
j=0 j + 5
converges or diverges.
Solution: √
Let a = j. It is obvious that a ≥ 0 for all j. It’s not so obvious that it’s
j j+5 j √
decreasing, but if you take the derivative of f(x) , you’ll see that f (x) < 0
x+5
for all x > 5. This implies that for j > 5, the series is decreasing. Finally,
√ j 1
lim = lim √ √ = 0
j + 5 j + 5/ j
√ √ √ 1
because lim5/ j = 0, so j +5/ j tends to ∞ and thus √j+5/ jtends to 0. So
all the conditions of the alternating series test are met, and therefore the series
converges. (It does not converge absolutely, but we didn’t ask for that.)
(d) Determine whether the series
X k lnk
(k + 1)
k=1
converges or diverges.
Solution:
It converges, which can be seen either by using the limit comparison test with
bk= k −3/2(see assignment solutions for details), or by arguing that
k lnk lnk
3 < 2 = ck.
(k + 1) k
P P
Then you can show that ckconverges using the integral test, and so k l3 k
k=1(k+1)
converges by the comparison test. (Note that all the terms in all the series here
are ≥ 0 as long as k ≥ 3, so the comparison test, limit comparison test, and
integral test can be applied.)
2
2. (a) Find the Maclaurin series of f(x) = ln(1 + x ), and determine its radius and
interval of convergence. Also discuss convergence at the endpoints of the interval.
2 Solution: Take the derivative to ﬁnd f (x) =2x2. Using the formula for the
1+x
geometric series,
∞ ∞
2x 2x X 2 n X n 2n+1
2 = 2 = 2x (−x ) = 2 (−1) x ,
1 + x 1 − (−x ) 0 0
provided that |x| < 1. You know that power series can be integrated term by
term, so
X∞ x2n+2 X∞ x2n+2
ln(1 + x ) = C + 2 (−1)n = C + (−1)n .
0 2n + 2 0 n + 1
2
Since ln(1 + 0 ) = ln1 = 0, you get that C must be 0 (because all the terms of
the series are 0 if x = 0). Thus
X x2n+2
ln(1 + x ) = (−1) n .
n + 1
0
Alternatively, if you happened to know the series for ln(1−t) by heart, you could
have used that and substituted −x for t to get the same answer. Now the ratio
test tells you that the radius of convergence is 1. You can also argue this by saying
that the radius of convergence doesn’t change when you integrate or diﬀerentiate
term by term, and you know it’s 1 for the geometric series. Finally, when x = 1
or x = −1, you get the convergent alternating series
∞ n
X (−1)
,
0 n + 1
so the interval of convergence is the closed interval [−1,1].
(b) State the Maclaurin series for f(x) = sinx. Also state its radius of covergence.
(“state” means that no justiﬁcation is required.)
Solution: X 2n+1
n x
sinx = (−1) (2n + 1)!
0
which converges for all values of x (i.e. R = ∞).
(c) Approximate the value of Z
0.5 2
sin(t )dt
0
correct to 6 decimal places (i.e. |error| < 0.000001). Justify your answer. (You
(0.5)
may use the fact that ≤ 0.0000004.)
1320
Solution:
Using the series for sint, we get that
∞ ∞
2 X n (t )n+1 X n t4n+2
sin(t ) = (−1) = (−1) .
0 (2n + 1)! 0 (2n + 1)!
3 Integrating term by term, you get that
Z 0.5 X∞ Z 0.5 t4n+2 X∞ t4n+3
sin(t )dt = (−1) n dt = [ (−1) n ]0.5
0 0 (2n + 1)! (4n + 3)(2n + 1)! 0
0 0
which equals
X∞ 4n+3 3 7

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