Math 347
Worksheet: Epsilonics, II: Infinite Series SOLUTIONS
A.J. Hildebrand
Worksheet: Epsilonics, II: Infinite Series SOLUTIONS
Practice Problems
Below are some problems to practice proof-writing skills in the context of infinite series. All of these proofs should be done
rigrously, using the “official” definitions of series convergence and results such as Cauchy’s Criterion. None of these proofs is
particularly difficult; try to master them all! For additional practice problems, especially of the “prove or find counterexample”
variety, see this week’s homework assignment.
P∞
A common mistake is to try
byP
directly manipulating infinite sums k=1 ak . E.g., trying
P∞to “prove” suchPresults
P∞to prove
∞
∞
the
sum
property
by
writing
(a
+
b
)
=
a
+
b
,
or
the
Absolute
Convergence
Test
by
writing
|
k
k
k
k
k=1
k=1
k=1
k=1 ak | ≤
P∞
k=1 |ak |, would be totally wrong and would completely miss the point of having precise mathematical definitions of convergence
and divergence of series.
P∞ (With this sort of manipulation one could “prove” all sorts of nonsensical results.)
An infinite sum
k=1 ak has a priori no meaning except as a formal expression. You should
Pn NEVER work with
infinite sums in your proof. P
Always work with finite sums, such as the partial sums
k=1 ak , or the sums of
n
finite “chunks” of the series, k=m+1 ak , that arise in Cauchy’s criterion.
1. Convergence/divergence of particular series:
(a) Geometric series, convergent
P∞ case: Using the definition of convergence of an infinite series, prove that if |r| < 1,
then the geometric series k=0 rk converges with sum 1/(1 − r). (Hint: Use the formula for the sum of a finite
geometric series.)
Proof: Let sn =
Pn
k=0
rk denote the n-th partial sum of the series. We seek to show that lim sn = 1/(1 − r).
n→∞
By the formula for the sum of a finite geometric series, we have
(1)
sn =
n
X
rk =
k=0
1
r
1 − rn+1
=
−
rn .
1−r
1−r
1−r
Since |r| < 1, we have lim rn = 0 (for a formal proof see below). Using the properties of limits, it follows that
n→∞
1
r
−
rn
(by formula (1))
n→∞
1−r
1−r
1
r
= lim
−
lim rn (by algebraic properties of limits)
n→∞ 1 − r
1 − r n→∞
1
= lim
− 0 (since (∗) lim rn = 0; see below)
n→∞ 1 − r
n→∞
1
=
(since lim c = c; see Problem 1(a) on the Epsilonics Worksheet).
n→∞
1−r
lim sn = lim
n→∞
Thus, the P
partial sums sn converge to 1/(1 − r). By the definition of convergence of an infinite series, this shows that
k
the series ∞
k=0 r converges, with sum 1/(1 − r).
Proof of (∗): Let an = |rn | = |r|n . Since |r| < 1, the sequence {an } is bounded and monotonically decreasing.
By the Monotone Convergence Theorem, it therefore has a limit, say L = lim |rn |. To show that L = 0, we
n→∞
apply the properties of limits: L = lim |rn+1 | = lim |r| · |rn | = |r| lim |rn | = |r|L, or L(1 − |r|) = 0. Since
n→∞
n→∞
n→∞
|r| < 1, this implies L = 0, as desired.
(b) Geometric series, divergent case: Prove that if |r| ≥ 1, then the series
appropriate convergence test.)
P∞
k=0
rk diverges. (Hint: Use an
Proof:PIn the case |r| ≥ 1 we have |rn | = |r|n ≥ 1, so rn does not have limit 0 as n → ∞. By the n-th term test, the
k
series ∞
k=0 r therefore diverges.
P∞
(c) Divergence of harmonic series: Prove that the harmonic series k=1 k1 diverges. (Hint: Show that Cauchy’s
Criterium for Series with is not satisfied, i.e., prove that it satisfies the negation of Cauchy’s Criterion.)
Proof: Let = 1/2. Then, for any N ∈ N, we have
2N
X
k=N +1
1
1
1
≥N·
= .
k
2N
2
Thus, the harmonic series does not satisfy the Cauchy Criterion (with the choices = 1/2 and m = N, n = 2N ), and
hence diverges.
2. General Properties of Series.
Math 347
Worksheet: Epsilonics, II: Infinite Series SOLUTIONS
A.J. Hildebrand
(a) Proof of Cauchy Criterion for Series: Prove the Cauchy Criterion for series, using the Cauchy Criterion for
sequences.
Proof: By definition, convergence of an infinite series is equivalent to convergence of the sequence of its partial sums
sn ; by the Cauchy Criterion for sequences, the sequence {sn } converges if and only if
(∀ > 0)(∃N ∈ N)(∀m, n ≥ N )[|sn − sm | < ].
(1)
Next, observe that we may in (1) assume that n ≥ m since interchanging m and n does not affect the quantity
|sn − sm |; in addition, we may assume n 6= m since when n = m, |sn − sn | = 0, so the inequality |sn − sn | < is
satisfied for any > 0. Thus, (1) is equivalent to
(∀ > 0)(∃N ∈ N)(∀n > m ≥ N )[|sn − sm | < ].
P
Now, by the definition of sn as the n-th partial sum of the series ∞
k=1 ak , we have (under the condition n > m)
(2)
sn − sm =
(3)
n
X
ak −
k=1
m
X
k=1
ak =
n
X
ak .
k=m+1
Substituting this into (2), we can rewrite (2) as
" n
#
X
(∀ > 0)(∃N ∈ N)(∀n > m ≥ N ) ak < .
(4)
k=m+1
The latter is the Cauchy Criterion for Infinite Series. To summarize, we have the following chain of equivalences:
∞
X
ak converges ⇐⇒ {sn } converges
k=1
⇐⇒ {sn } satisfies Cauchy Criterion for sequences, i.e., (1)
⇐⇒ {sn } satisfies (2)
⇐⇒
⇐⇒
∞
X
k=1
∞
X
ak satisfies (4)
ak satisfies Cauchy Criterion for infinite series
k=1
(b) Proof of Absolute Convergence Test: Prove the Absolute Convergence Test using the Comparison Test.
P
P∞
Proof: The Absolute Convergence Test says that if ∞
k=1 |ak | converges, then so does
k=1 ak . This follows immediately from the Comparison Test using bk = |ak | as the terms in the comparison series.
(c) Series of nonnegative terms: Prove that if a series has (i) only nonnegative terms and (ii) bounded partial sums,
then it converges. (Hint: What does the “nonnegative terms” condition mean in terms of the partial sums sn ?)
Pn
Proof: Since sn+1 − sn = an+1 , the condition ak ≥ 0 for all k ∈ N implies that the partial sums sn =
k=1 ak
satisfy sn+1 ≥ sn for all n ∈ N, i.e., they form a monotone sequence. Since we are also given that the partial sums
are bounded, the MonotonePConvergence Theorem applies and yields that the sequence {sn } converges. By definition
this means that the series ∞
k=1 ak converges.
P∞
(d) Proof of the n-th term test: The n-th term test says that if k=1 ak converges, then limn→∞ an = 0. Give a
careful proof of the this test, by two methods: (i) using the definition of convergence of infinite series and algebraic
properties of limits of sequences; (ii) using the Cauchy Criterion for Series.
Proof: Assume
(1)
P∞
k=1
ak converges. Let > 0 be given. By Cauchy’s Criterion there exists an N1 ∈ N such that
n
X
ak < for all n, m ∈ N with n > m ≥ N1 .
k=m+1
In particular, applying this with n = m + 1, (1) yields
(2)
|an | < for all n ∈ N with n > N1 .
Now let N = N1 + 1. Then n > N1 is equivalent to n ≥ N , so (2) says that n ≥ N implies |an − 0| < . By the
definition of a limit, this proves limn→0 an = 0.
P∞
P∞
(e) Changing finitely many terms in a series does not affect convergence: Suppose k=1 ak and k=1 bk are
series whose terms differ in at most finitely many places; i.e.,Pthere exists an N ∈ N such thatP
bk = ak for all k ≥ N .
∞
∞
Using the definition of convergence of a series, prove that if k=1 ak converges, then so does k=1 bk .
Proof: This follows by comparing
the Cauchy
Pn
Pn Criterion for the two series. Let N0 be such that ak = bk for all
k ≥ N0 . Then the sums
k=m+1 ak and
k=m+1 bk , arising in this criterion are identical provided we choose N
(which is at
our
disposal)
large
enough
so
that
N ≥ N0 . (Specifically, given
P
P∞ > 0, let Na be an N that “works” for
the series ∞
a
.
Then
N
=
max(N
,
N
)
will
“work”
for
the
series
a
0
k
b
k=1
k=1 bk .)