MATH 101 Quiz #5 (v.A2)
Last Name:
Friday, March 18
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Does the series
∞
X
n=2
n2
√ converge? (Answer “yes” or “no” in the box.)
3n2 + n
Answer: No
Solution: We have
n2
1
1
√
√
6= 0,
=
lim
=
n→∞ 3n2 +
3
n n→∞ 3 + 1/(n n)
lim
so the series diverges by the Test for Divergence.
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
2. 2 marks Relate the number 0.53̄ = 0.5333333 . . . to the sum of a geometric series, and use
that to represent it as a rational number (a fraction or combination of fractions, with no
decimals). A calculator-ready answer is acceptable.
Answer:
3
3
Solution: The number is 0.5 + 100
+ 1000
+ 1034 + · · · = 12 + 1032
series sums to
3
1
3
1
=
=
,
1
102 1 − 10
10(10 − 1)
30
so the fraction is
8
15
P∞
n=0
10−n . The geometric
1
1
8
+
= .
2 30
15
Marking scheme:
• 1 mark for writing the geometric series (either with
• 1 mark for getting the right answer
P
or · · · notation)
3. 2 marks Show that the series
∞
X
n=2
3
converges.
n(log n)5/4
3
. Then f (x) is positive and decreasing, so that the sum
x(log x)5/4
Z ∞
∞
X
f (n) and the integral
f (x) dx either both converge or both diverge, by the Integral
Solution: Let f (x) =
2
2
Test. For the integral, we use the substitution u = log x, du =
Z ∞
Z ∞
3 dx
3 du
=
,
5/4
x(log x)5/4
2
log 2 u
dx
x
to get
which converges by the p-test (5/4 > 1).
Marking scheme:
• 1 point for correctly using the Integral Test. (For this quiz, students can earn this mark
even if they don’t state that f (x) is positive and decreasing; however, on the final exam,
students must show that they know these hypotheses are important.)
• 1 point for showing the integral converges (citing the p-test is OK)
Long answer question—you must show your work
4. 5 marks Find the solution to the differential equation
yy 0
1
= that satisfies y(0) = 3.
x
e − 2x
y
Solve completely for y as a function of x.
Solution: Cross-multiplying, we rewrite the equation as
dy
= ex − 2x
dx
y 2 dy = (ex − 2x) dx.
y2
Integrating both sides, we find
1 3
y = ex − x2 + C.
3
Setting x = 0 and y = 3, we find C = 8; therefore the solution is
1 3
y = ex − x 2 + 8
3
y = (3ex − 3x2 + 24)1/3 .
Marking scheme:
• 1 mark for separating the variables
• 2 marks for integrating both sides of the equation
• 1 mark for finding the value of the constant
• 1 mark for solving for y