Download Series Convergence Tests Math 122 Calculus III and more Study notes Calculus in PDF only on Docsity!
Series Convergence Tests
Math 122 Calculus III
D Joyce, Fall 2012
Some series converge, some diverge.
Geometric series. We’ve already looked at these. We know when a geometric series
converges and what it converges to. A geometric series
∑^ ∞
n=
arn^ converges when its ratio r lies
in the interval (− 1 , 1), and, when it does, it converges to the sum
a 1 − r
The harmonic series. The standard harmonic series
∑^ ∞
n=
n
diverges to ∞. Even though
its terms 1, 12 , 13 ,... approach 0, the partial sums Sn approach infinity, so the series diverges.
The main questions for a series.
Question 1: given a series does it converge or diverge? Question 2: if it converges, what does it converge to?
There are several tests that help with the first question and we’ll look at those now.
The term test. The only series that can converge are those whose terms approach 0. That
is, if
∑^ ∞
k=
ak converges, then ak → 0.
Here’s why. If the series converges, then the limit of the sequence of its partial sums
approaches the sum S, that is, Sn → S where Sn is the nth^ partial sum Sn =
∑^ n
k=
ak. Then
lim n→∞ an = lim n→∞ (Sn − Sn− 1 ) = lim n→∞ Sn − lim n→∞ Sn− 1 = S − S = 0.
The contrapositive of that statement gives a test which can tell us that some series diverge.
Theorem 1 (The term test). If the terms of the series don’t converge to 0, then the series diverges.
Note, however, the terms converging to 0 doesn’t imply the series converges, as the har- monic series gives a counterexample to that. The term test can be used to show that the following series don’t converge ∑ (^) n n + 1
∑ (^) n! n^2
(−1)n^
n 2 n + 1
because their terms do not approach 0. The rest of the tests in this note deal with positive series, that is, a series none of whose terms are negative. Note that the only way a positive series can diverge is if it diverges to infinity, that is, its partial sums approach infinity.
The comparison test. Essentially, a positive series with smaller terms sums to a smaller number than a series with larger terms.
Suppose that
∑^ ∞
k=
ak and
∑^ ∞
k=
bk are two positive series and every term of the first is less
than or equal to the corresponding term of the second, that is, an ≤ bn for all n. If that’s the
case, we’ll say the second series dominates the first series. Then each partial sum Sn =
∑^ n
k=
ak
of the first series is less than or equal to the corresponding partial sum Tn =
∑^ n
k=
bk of the
second series. Furthermore, since these are both positive series, so both sequences {Sn}∞ n= and {Tn}∞ n=1 of partial sums are increasing sequences. Suppose the second sequence converges to a number T , that is, Tn → T. Then T bounds the sequence {Tn}∞ n=1, so it also bounds the smaller sequence {Sn}∞ n=1. Since an increasing bounded sequence has a limit, therefore {Sn}∞ n=1 has a limit, S, and S ≤ T. Thus, we’ve proven the following theorem. The second statement is the contrapositive of the first, so it’s also true.
Theorem 2 (The comparison test). Suppose that one positive series is dominated by another. If the second converges, then so does the first. If the first diverges to infinity, then so does the second.
You can think of this theorem as simply saying that
If an ≤ bn for each n, then
an ≤
bn
Example 3. Any series dominated by a positive convergent geometric series converges. For
instance, we’ll show
∑^ ∞
n=
n!
converges since it’s dominated by the convergent geometric series
∑^ ∞
n=
2 n^
. All we need to do is show that
n!
2 n^
for large n. But for n ≥ 4, 2n^ ≤ n!. Thus
∑^ ∞
n=
n!
is dominated by a convergent geometric series, and, so, it’s also a convergent series.
Since the series
∑^ ∞
n=
n!
only has finitely many more terms, it also converges.
Note that whether a series converges or diverges doesn’t at all depend on the first few terms. It only depends the rest of them, the “tail” of the series. For that reason, when we’re only interested in convergence, we’ll leave usually abbreviate our sigma notation and
say
n!
converges.
p-series Series of the form
np^
, where p is a constant power, are called p-series. When
p = 1, the p-series is the harmonic series which we know diverges. When p = 2, we have the convergent series mentioned in the example above. By use of the integral test, you can determine which p-series converge.
Theorem 7 (p-series). A p-series
np^
converges if and only if p > 1.
Proof. If p ≤ 1, the series diverges by comparing it with the harmonic series which we already know diverges. Now suppose that p > 1. The function f (x) = 1/xp^ is a decreasing function, so to determine the convergence of the series we’ll detemine the convergence of the corresponding integral.
∫ (^) ∞
1
xp^
(p − 1)xp−^1
∞ 1
p − 1
Since the integral converges, so does the series. q.e.d.
Some example divergent p-series are
n
and
n
. Some convergent ones are
n^2
n
n
, and
n^1.^001
The limit comparison test. This test is an improvement on the comparison test. It incorporates the fact that a series converges if and only if a constant multiple of it converges (provided that constant is not 0, of course). So long as you can compare a multiple of one series to another, that’s enough to do a comparison.
Theorem 8 (Limit comparison test). Given two positive series
an and
bn where the ratio of their terms an/bn approaches a positive number, then they either both converge or diverge, that is,
an ∼
bn.
Proof. Suppose that an/bn → L, a positive number. Let = L/2. Then for large n, |an/bn − L| < L/2. Hence, L 2 bn ≤ an ≤ 32 L bn. Now, if
an converges, then by the comparison test, so does
∑ L
2 bn^ converge, hence^
bn converges. On the other hand, if
∑ bn, converges, so does 3 L 2 bn, and again by the comparison test,^
an converges. q.e.d.
One of the applications of the limit comparison test is that it allows us to ignore small
terms. Consider the series
an =
∑ (^3) n^2 + 2n + 1 n^3 + 1
. We can replace this series by
bn =
∑ (^) n^2
n^3
n
because
an bn
3 n^2 + 2n + 1 n^3 + 1
/n 2
n^3
3 n^2 + 2n + 1 n^2
n^3 n^3 + 1
But
bn is the harmonic series, which diverges. Therefore our original series
an also diverges.
The root test. The root test doesn’t have a lot of applications, but I’m including it here since it’s one of the standard tests. For the root test, you look at the limit of the nth^ root of the nth^ term.
Theorem 9 (The root test). If lim n→∞ a^1 n/n = L, and if L < 1 then the series converges, but if
L > 1 the series diverges.
For the root test, if L = 1, then the test is inconclusive, so you have to use some other test.
Example 10. The root test is especially useful when the nth^ term already has a nth^ power
in it. Consider the series
(ln n)n^
. Here, a^1 n/n =
(ln n)n
) 1 /n
ln n
→ 0. So that series
converges.
Here’s the proof for the root test in the case that L < 1. The case L > 1 is analogous. We’ll show
an converges by comparing it to a larger convergent geometric series. Let r be
a number between L and 1. Since a^1 n/n → L, therefore for sufficiently large n, a^1 n/n < r, so an < rn. But the geometric series
rn^ converges, so
an also converges. q.e.d.
The ratio test. We won’t use the root test a lot, but the ratio test is very important, and we’ll use a version of it soon on every power series we analyze. The statement of it is similar
to that of the root test. For the ratio test, you look at the limit of the ratio
an+ an
of adjacent
terms.
Theorem 11 (The ratio test). If lim n→∞
ak+ ak
= L, and if L < 1 then the series converges, but
if L > 1 the series diverges.
For the ratio test, as it was for the root test, if L = 1, then the test is inconclusive, so you have to use some other test.
Example 12. Consider the series
∑ 7 n n!
. The nth^ term is an =
7 n n!
so the next term is
an+1 =
7 n+ (n + 1)!
. For this ratio test, we’ll examine the ratio
an+ an
and find its limit.
an+
an =
7 n+ (n + 1)!
/ 7 n
n!
7 n+ 7 n
n! (n + 1)!
n + 1
Since the limit L = 0 is less than 1, this series converges.
The proof of the first case of the ratio test depends on comparing it to a larger convergent geometric series like we did for the root test. Again, we’ll do just the case L < 1 here since the case L > 1 has an analogous proof.
Let r be a number between L and 1. Since
an+ an
, therefore for sufficiently large n, say
n ≥ N , we have
an+ an
< r, and so an+1 < anr. A quick inductive argument shows that
an < aN rn−N^. That says the tail of our series is dominated by a convergent geometric
series,
∑^ ∞
n=N
an <
∑^ ∞
n=N
aN rn−N^. Therefore, by the comparison test, the original series also
converges. q.e.d.
