James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

728 Chapter 11 Infinite Sequences and Series
It is important to keep in mind the
distinction between a sequence and a
series. A sequence is a list of numbers,
whereas a series is a sum. With every
series o a n there are associated two
sequences: the sequence ha nj of terms
and the sequence hs nj of partial sums.
Standard Series for Use
with the Comparison Test
Proof
(i) Let
s n − o n
a i
i−1
t n − o n
b i
i−1
t − ò b n
n−1
Since both series have positive terms, the sequences hs n j and ht n j are increasing
ss n11 − s n 1 a n11 > s n d. Also t n l t, so t n < t for all n. Since a i < b i , we have s n < t n .
Thus s n < t for all n. This means that hs n j is increasing and bounded above and therefore
converges by the Monotonic Sequence Theorem. Thus o a n converges.
(ii) If o b n is divergent, then t n l ` (since ht n j is increasing). But a i > b i so s n > t n .
Thus s n l `. Therefore o a n diverges.
n
In using the Comparison Test we must, of course, have some known series o b n for
the purpose of comparison. Most of the time we use one of these series:
• A p -series 3o 1yn p con verges if p . 1 and diverges if p < 1; see (11.3.1)4
• A geometric series 3o ar n21 converges if | r | , 1 and diverges if | r | > 1;
see (11.2.4)4
ExamplE 1 Determine whether the series ò
n−1
5
converges or diverges.
2n 2 1 4n 1 3
SOLUtion For large n the dominant term in the denominator is 2n 2 , so we compare the
given series with the series o 5ys2n 2 d. Observe that
5
2n 2 1 4n 1 3 , 5
2n 2
because the left side has a bigger denominator. (In the notation of the Comparison Test,
a n is the left side and b n is the right side.) We know that
ò
n−1
5
2n 2
− 5 2 ò n−1
is convergent because it’s a constant times a p-series with p − 2 . 1. Therefore
ò
n−1
1
n 2
5
2n 2 1 4n 1 3
is convergent by part (i) of the Comparison Test.
n
note 1 Although the condition a n < b n or a n > b n in the Comparison Test is given
for all n, we need verify only that it holds for n > N, where N is some fixed integer,
because the convergence of a series is not affected by a finite number of terms. This is
illustrated in the next example.
ln k
Example 2 Test the series ò for convergence or divergence.
k−1 k
SOLUtion We used the Integral Test to test this series in Example 11.3.4, but we can
also test it by comparing it with the harmonic series. Observe that ln k . 1 for k > 3
and so
ln k
. 1 k > 3
k k
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