MATH1120 Calculus II
Quiz 3 Solution
November 21, 2013
Name:
Section:
Time allowed: 20 ± ε minutes
Full score: 40+10
Instruction: Calculators, notes, textbooks and computer-printed cheat sheets are not allowed.
You may use an A4-size handwritten cheat sheet. Please show your work clearly to receive full
credit. Good luck!
1. (5 pts) Write down your expected score.
(Your expected score)
2. (5 pts) Suppose lim an = 0. Show, by Squeeze Theorem, that lim (−1)n an = 0.
n→∞
n→∞
Solution: Note that −an ≤ (−1)n an ≤ an , and lim an = lim −an = 0. By Squeeze
n→∞
Theorem, lim (−1)n an = 0.
n→∞
n→∞
3. Determine the convergence of the following series. If the series converges, determine
whether it converges absolutely or conditionally. State which convergence tests you use,
how to use them, and the series with which you compare the given series wherever appropriate. Remember to point out the assumptions of the convergence tests (if any) the given
series and the comparison series satisfy.
∞
X
1
(a) (5 pts)
sin
(Hint: compare with one p-series)
n
n=1
∞
X1
sin 1
1
> 0 for all n, and lim 1 n = 1 > 0. Since
diverges
n→∞
n
n
n
n=1
by the p-series test, so does the given series by the Limit Comparison Test.
∞
X
cos n
(b) (5 pts)
n2
n=1
∞
cos n 1 X 1
Solution: Note that 0 ≤ 2 ≤ 2 .
converges by the p-series test. By the
n
n
n2
n=1
∞
∞ X
X
cos n
cos n Direct Comparison Test,
converges absolutely. By
2 converges. So
n
n2
n=1
n=1
∞
X
cos n
the Absolute Convergence Test,
converges.
n2
Solution: Note that sin
n=1
(c) (10 pts)
∞
X
(−1)n √
n=1
1
n2 + 3
Solution: Note that un = √
1
, un > 0 for all n, un > un+1 for all n, and
+3
lim un = 0. By the Alternating Series Test, the given series converges. On the
n2
n→∞
1
∞
X
3
1
other hand, because
= lim 1 + 2 = 1 and
diverges by the
n→∞
n
n
n=1
X
∞
1
1
n
=
√
√
p-series test, it follows that
(−1)
diverges by the Limit
n2 + 3
n2 + 3
n=1
n=1
Comparison Test. Therefore the given series converges conditionally.
∞ X
1 n n
4. (10 pts) Find the radius and interval of convergence of the power series
1+
x .
n
1
n
lim
n→∞ √ 1
n2 +3
∞ X
r
n=1
Solution: We use the Root Test.
1
1 n n n
1+
lim
x <1
n→∞ n
1
|x| < 1
lim 1 +
n→∞
n
|x| < 1
So the radius of convergence is 1, and the power series converges absolutely forx ∈ (−1,
1).
1 n
=
When x = ±1, the power series diverges by the n-th term test, as lim ± 1 +
n→∞
n
±e 6= 0. So the interval of convergence is (−1, 1).
5. Bonus Problem (10 pts) Find a closed-form expression for the power series x + 2x2 + 3x3 +
· · · + nxn + · · · for |x| < 1. (Hint: divide by x and then integrate)
Solution: The given power series converges absolutely for |x| < 1. Suppose it converges to
f (x) for |x| < 1.
x + 2x2 + 3x3 + · · · + nxn + · · · = f (x)
f (x)
1 + 2x + 3x2 + · · · + nxn−1 + · · · =
Z x
Z x
Z x
Z xx
Z x
f (t)
dt +
2tdt +
3t2 dt + · · · +
ntn−1 dt + · · · =
dt (Integration term by term)
t
0
0
0
0
0
Z x
f (t)
2
3
n
dt
x + x + x + ··· + x + ··· =
t
Z0 x
x
f (t)
=
dt
1−x
t
0
d x
f (x)
=
(By FTC)
dx 1 − x
x
x
f (x) =
(1 − x)2
2