# Very short answer question

```MATH 101 Quiz #5 (v.A1)
Last Name:
Friday, March 18
First Name:
Student-No:
Section:
Very short answer question
1. 1 mark Does the series
1
X
n=2
n
p converge? (Answer “yes” or “no” in the box.)
2n + n
Solution: We have
n
1
1
p = lim
p = 6= 0,
n!1 2n +
2
n n!1 2 + 1/ 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 0.86̄ = 0.866666 . . . 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).
6
6
Solution: The number is 0.8 + 100
+ 1000
+ 1064 + &middot; &middot; &middot; = 45 + 1062
series sums to
6
1
6
1
= ,
1 =
2
10 1 10
10(10 1)
15
so the fraction is
13
15
P1
n=0
10
n
. The geometric
4
1
13
+
= .
5 15
15
Marking scheme:
• 1 mark for writing the geometric series (either with
• 1 mark for getting the right answer
P
or &middot; &middot; &middot; notation)
3. 2 marks Show that the series
1
X
n=3
5
converges.
n(log n)3/2
5
. Then f (x) is positive and decreasing, so that the sum
x(log x)3/2
Z
1
1
X
f (n) and the integral
f (x) dx either both converge or both diverge, by the Integral
Solution: Let f (x) =
3
3
Test. For the integral, we use the substitution u = log x, du =
Z 1
Z 1
5 dx
5 du
=
,
3/2
x(log x)3/2
3
log 3 u
dx
x
to get
which converges by the p-test (3/2 &gt; 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 di↵erential equation
Solve completely for y as a function of x.
yy 0
ex + e
x
=
Solution: Cross-multiplying, we rewrite the equation as
dy
= ex + e x
dx
y 2 dy = (ex + e x ) dx.
y2
Integrating both sides, we find
1 3
y = ex
3
Setting x = 0 and y = 5, we find C =
125
;
3
e
x
+ C.
therefore the solution is
1 3
125
y = ex e x +
3
3
x
x
y = (3e
3e + 125)1/3 .
Marking scheme:
1
that satisfies y(0) = 5.
y
• 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
```