MATH 152 - Recitation Quiz 7 - Spring 2015
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1. The terms of a series are dened recursively by the equations
a1 = 2, an+1 =
Determine whether the series
P
5n + 1
an .
4n + 3
an converges or diverges.
Method 1:
Note that
5n+1 a 4n+3 n 5n + 1 5
an+1 = > 1.
= lim lim lim
= n→∞
n→∞
n→∞ an an
4n + 3 4
Therefore, by the Ratio Test, the series diverges.
Method 2:
Note that if n ≥ 2, then 5n − 4n ≥ 3 − 1 =⇒ 5n + 1 ≥ 4n + 3. Therefore, if n ≥ 2,
5n + 1
5n + 1
≥ 1 =⇒
an ≥ an =⇒ an+1 ≥ an .
4n + 3
4n + 3
Note also that
a2 =
6
12
a1 =
.
7
7
Therefore, every term of the sequence (an ) is greater than or equal to 2. Therefore, either
lim an = ∞ or lim an ≥
n→∞
n→∞
Therefore,
lim an 6= 0.
n→∞
Thus, by the Test for Divergence,
P
an diverges.
1
12
.
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2. Determine whether the following series diverges or converges.
∞
X
(n + 1) 5n
n32n
n=1
Method 1:
Note that
(n + 2) 5n+1
n32n 5
= < 1.
lim
·
n→∞ (n + 1) 32n+2 (n + 1) 5n 9
Therefore, by the Ratio Test, the series converges.
Method 2:
Note that
r
lim
n→∞
n
(n + 1) 5n
= lim
n→∞
n32n
r
n
n+1
·
n
s n
5
5
n
= < 1.
9
9
Therefore, by the Root Test, the series converges.
Method 3:
Note that
n
∞
∞
X
X
(n + 1) 5n
(n + 1)
5
=
·
.
2n
n3
n
9
n=1
n=1
Also,
h
lim
n→∞
Recall that since 59 < 1, that ∞
n=1
Test, the original series converges.
P
5 n
9
(n+1)
n
· 95
5 n
n i
= 1.
9
is a convergent geometric series. Therefore by the Limit Comparison
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3. Find the radius of convergence and interval of convergence of the following series.
∞
n
X
(x − 4)
n5n
n=1
First, note that the center of the interval of convergence is 4. We now use the ratio test to determine the
radius of convergence of the series.
(x − 4)n+1
n 1
n5n · |x − 4| = 1 |x − 4| .
=
lim
·
lim n
n+1
n→∞ (n + 1) 5
5
(x − 4) n→∞ n + 1 5
Since the ratio test applies only if the limit above is less than 1, and
1
|x − 4| < 1 =⇒ |x − 4| < 5,
5
the radius of convergence of the series is 5.
Hence, if x ∈ (4 − 5, 4 + 5) = (−1, 9) , then the series converges. However, we have not found the interval
of convergence yet, because we have to determine the behavior of the series at the endpoints of the interval.
First we check what happens if x = −1 :
n X
∞
∞
∞
n
n
X
X
(x − 4)
1
−5
(−1)
=
·
=
.
n5n
n
5
n
n=1
n=1
n=1
This is the alternating harmonic series, which converges by the alternating series test. Therefore, the series
converges when x = −1. Now we check what happens if x = 9 :
n X
∞
∞
∞
n
X
X
(x − 4)
1
5
1
=
·
=
.
n
n5
n
5
n
n=1
n=1
n=1
This is the harmonic series, which diverges by the p-test. Therefore, the series diverges when x = 9. This
gives us the interval of convergence [−1, 9) .
3