MATH 107-153 Recitation 22
JD Nir
Avery 230 • Office Hours: W 4-5 R 11-2
[email protected]
www.math.unl.edu/∼jnir2/107-153.html
November 17, 2015
1: Determine if the following series converge absolutely, conditionally or diverge.
(a)
∞
X
n(−2/3)n
n=1
First, we rewrite this sum to look more like the examples we are used to:
∞
X
n(−2/3)n =
n=1
∞
X
(−1)n n2n
n=1
3n
We note that this is an alternating series. We could use the Alternating Series Test, but it is not
immediately obvious that the series is decreasing or that it’s limit is zero. This method does work,
but it turns out to be way more work is necessary.
Instead, we use the Ratio Test:
(−1)n+1 (n + 1)2n+1 an+1 3n+1
= lim lim n
n
n→∞
n→∞ (−1) n2
an
3n
n
(−1)n+1 (n + 1)2n+1
3
Note that from now on we skip directly to this step.
·
= lim n→∞
3n+1
(−1)n n2n (−1)n+1 n + 1 2n+1 3n = lim ·
· n · n+1 n→∞
(−1)n
n
2
3
1
n+1
= lim −1 ·
· 2 · n→∞
n
3
2n + 2
= lim
n→∞
3n
2 + n2
2
= lim
= <1
n→∞
3
3
Because the limit of the ratio converges to 23 < 1, the series converges by the ratio test. Furthermore, it converges absolutely because the ratio test considers the absolute value of the limit, so
we would find the same result if we considered the absolute value of the series.
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MATH 107-153 Recitation 22
∞
X
n!
3n
n=0
This looks like a perfect series to use the Ratio Test!
(b)
an+1 (n + 1)! 3n = lim lim · n→∞ an n→∞ 3n+1
n!
(n + 1)! 3n · n+1 = lim n→∞
n!
3
1
= lim (n + 1) · n→∞
3
=∞
Because the limit of the ratio diverges, the series diverges by the ratio test.
∞
X
(−1)n
√
(c)
n
n=1
Because roots of n do not cancel, the ratio test does not look to be particularly useful here. Instead,
we apply the Alternating Series Test.
√
√
1
because n < n + 1.
We first check that |an | = bn > bn+1 = |an+1 |. We note √1n > √n+1
Next we check that lim an = 0. This is also very easy to check.
n→∞
Thus, because the bn are decreasing and the limit of an approaches 0, the series converges by the
alternating series test.
∞
X
1
Now consider what happens when we take the absolute value of the series. We know
1 din
n=1 2
verges by the p-series test. Thus, we know that the original series converges conditionally.
∞
X
(−1)n (2n2 − 1)
(d)
3n4 − 2
n=1
Using the ratio test would require expanding (n + 1)4 (why?) so instead we use the Alternating
Series Test.
2x2 −1
0
First we show the function is decreasing. We do this by allowing f (x) = 3x
4 −2 and finding f (x):
d (2x2 − 1)
(3x4 − 2)(4x) − (2x2 − 1)(12x3 )
=
dx 3x4 − 2
(3x4 − 2)2
12x5 − 8x − 24x5 + 12x3
=
(3x4 − 2)2
−12x5 + 12x3 − 8x
=
(3x4 − 2)2
3
12x (1 − x2 ) − 8x
=
(3x4 − 2)2
−12x3 (x2 − 1) − 8x
=
(3x4 − 2)2
We see that the numerator is negative whenever x2 > 1 and the denominator is always positive. Thus, the derivative is negative and the function is decreasing. We conclude the sequence
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MATH 107-153 Recitation 22
(−1)n (2n2 −1)
3n4 −2
is also decreasing. Next, we check lim
n→∞
(−1)n (2n2 −1)
:
3n4 −2
(−1)n (2n2 − 1)
= lim
n→∞
n→∞
3n4 − 2
lim
=
2
1
− 4
2
n
n
2
3− 4
n
(−1)n
0
=0
3
Thus, because the sequence of absolute values is decreasing and the limit of the sequence is zero,
the series converges by the Alternating Series Test.
To determine whether it converges absolutely or conditionally, we check the limit behavior of
∞
∞
X
X
(2n2 − 1)
1
.
We
do
this
by
the
using
the
Limit
Comparison
Test
with
:
3n4 − 2
n2
n=1
n=1
2n2 − 1
4
(2n2 − 1)(n2 )
lim 3n − 2 = lim
1
n→∞
n→∞
3n4 − 2
n2
2n4 − n2
= lim
n→∞ 3n4 − 2
2 − n12
= lim
n→∞ 3 − 24
n
2
=
3
∞
∞
X
X
(2n2 − 1)
1
converges
by
the
p-series
test,
we
know
converges by the limit comn2
3n4 − 2
n=1
n=1
∞
X
(−1)n (2n2 − 1)
converges absolutely.
parison test, and thus
3n4 − 2
n=1
∞
X
nn
(e)
n!
Because
n=1
At first, the root test looks attractive because we raise something to the nth power. However,
this would require taking the nth root of n! which is difficult. Instead, we use the Ratio Test.
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MATH 107-153 Recitation 22
Hopefully you are beginning to understand how powerful this test is!
(n + 1)n+1 n! an+1 = lim lim
· n n→∞
n→∞ an (n + 1)!
n
n+1
(n + 1)
n! = lim ·
n→∞
nn
(n + 1)! (n + 1)n · (n + 1)
1
= lim ·
n
n→∞
n
n + 1
n + 1 n
= lim n→∞
n
n 1 = lim 1 +
n→∞
n =e>1
Because the limit of the ratio converges to e > 1, the series diverges by the Ratio Test.
2: For which values of x do the following converge?
(a)
∞ X
x n
k
n=0
This looks
like our alternate form of the geometric series. We know the series will converge if
x
r = k < 1. In other words, we need |x| < k or −k < x < k. We note that if x = k, we have
∞
∞
X
X
n
1 , which clearly diverges. Similarly, when x = −k, we have
(−1)n which does not converge.
n=0
n=0
Therefore, our interval of congergence does not include either endpoint. We say the series converges
on −k < x < k.
∞
X
(−1)n xn
(b)
n2
n=0
We use the ratio test, mostly because it has already been so useful.
(−1)n+1 xn+1
an+1 n2 = lim ·
lim
n→∞
n→∞ an (n + 1)2
(−1)n xn 2
(−1)n+1 xn+1
n
= lim · n ·
n
2
n→∞
(−1)
x
(n + 1) n2
= lim |x| · 2
n→∞
n + 2n + 1
1
= lim |x| ·
2
n→∞
1 + n + n12
= |x|
Now, we know that the ratio test converges if the limit of the ratios is less than one, so we have
∞
X
(−1)n
|x| < 1. Remember to check your boundaary values. For x = 1, we get
which converges
n2
n=0
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MATH 107-153 Recitation 22
by the Alternating Series Test (check this on your own). For x = −1, we get
by the p-series test. Thus, our radius of convergence is −1 ≤ x ≤ 1.
5
∞
X
1
which converges
n2
n=0