MATH 101 Quiz #6 (v.A1)
Last Name:
First Name:
Friday, April 1
Grade:
Student-No:
Section:
Very short answer question
P
1. 1 mark Suppose you wanted to use the Limit Comparison Test on the series ∞
n=0 an where
n
an
an = 23n+n
.
Write
down
a
sequence
{b
}
such
that
lim
exists
and
is
nonzero.
(You don’t
n
n→∞ bn
+1
have to carry out the Limit Comparison Test; just write the formula for the bn .)
Answer:
2n
3n
Solution: The most significant term in the numerator is 2n , and the most significant term
∞ n
X
2
n
.
in the denominator is 3 . Therefore we should compare this series to the series
3
n=0
Marking scheme: 1 for any correct answer in the box
Short answer questions—you must show your work
P
2
(−1)n−1
= π12 (you don’t have to show this). Find N so that
2. 2 marks It is known that ∞
n=1
n2
2
SN , the N th partial sum of the series, satisfies | π12 − SN | ≤ 10−6 . Be sure to say why your
method can be applied to this particular series.
Answer: 999
Solution: The sequence { n12 } decreases to zero, so by the Alternating Series error bound,
1
−6
the error is at most the first omitted term. We therefore need (N +1)
, which is
2 ≤ 10
3
equivalent to N + 1 ≥ 10 and N ≥ 999.
Marking scheme:
• 1 mark for saying that { n12 } is decreasing to zero (or decreasing and has limit 0, etc.)
• 1 mark for correctly justifying a value of N (any justified value is acceptable)
3. 2 marks Does the series
∞ √
X
n cos n
n=5
n2 − 1
converge conditionally, converge absolutely, or diverge?
Explain your answer.
√
Solution: We know that nn2cos(n)
≤
−1 √
n
,
n2 −1
since |cos(n)| ≤ 1 for all n.
Next, we have
√
lim
n→∞
√
n/(n2 − 1)
n · n3/2
n2
1
=
lim
= lim
=
lim
= 1,
3/2
2
2
n→∞ n − 1
n→∞ 1 − 1/n2
n→∞ n − 1
1/n
√
P
n
together with the p-series
so by the Limit Comparison Test the series ∞
n=5 n2 −1 converges
P∞ 1
P∞ √n cos(n) n=5 n3/2 . By the Comparison Test it follows that
n=5 n2 −1 converges, and hence
that the given series converges absolutely.
Marking scheme:
• 1 point for correctly using a convergence test (Limit Comparison Test or Comparison
Test) on a positive series
• 1 point for stating that the series converges absolutely
Long answer question—you must show your work
4. 5 marks Find the radius of convergence and interval of convergence of the series
n
∞
X
(−1)n x + 2
.
n+1
2
n=0
Solution: We have
(−1)n+1 /(n + 2) · ((x + 2)/2)n+1 an+1 = lim lim n→∞ n→∞ an (−1)n /(n + 1) · ((x + 2)/2)n n + 1 x + 2 |x + 2|
= lim
=
.
n→∞ n + 2 2 2
Therefore, by the Ratio Test, the series converges when
|x+2|
> 1. In particular, it converges when
2
|x+2|
2
< 1 and diverges when
|x + 2| < 2 ⇐⇒ −2 < x + 2 < 2 ⇐⇒ −4 < x < 0,
and the radius of convergence is R = 2. (Alternatively, one can set cn =
1
= , so that R = 1 = 2.)
compute L = limn→∞ cn+1
cn
2
L
(−1)n
(n+1)2n
and
P
(−1)n
Next, we consider the endpoints 0 and −4. At x = 0 the series is simply ∞
n=0 n+1 ,
which is an alternating series: the signs alternate, and the unsigned terms decrease to
zero. Therefore the series converges at x = 0 by the Alternating Series Test.
At x = −4 the series is
n X
∞
∞
∞
X
X
(−1)n −4 + 2
(−1)n
1
n
=
(−1) =
,
n+1
2
n+1
n+1
n=0
n=0
n=0
n
since (−1)n · (−1)n = (−1)2n = ((−1)2 ) = 1. This series diverges, either by comparison
or limit comparison with the harmonic series (the p-series with p = 1), or directlyPby the
1
Integral Test. (For that matter, it is exactly equal to the standard harmonic series ∞
n=1 n ,
re-indexed to start at n = 0.)
In summary, the interval of convergence is −4 < x ≤ 0, or simply (−4, 0].
Marking scheme:
• 1 mark for using the Ratio Test
• 1 mark for getting R = 2 (radius of convergence)
• 1 mark for showing convergence at x = 0 (any valid way)
• 1 mark for showing divergence at x = −4 (any valid way)
• 1 mark for writing the interval of convergence (in either form)