Math 104: Introduction to Analysis Summer 2019
Lecture 13: Alternating Series, Continuity
Lecturer: Michael Christianson 17 July Aditya Sengupta
Proof. (sketch of a proof of the Alternating Series Test) Because (an) is a decreasing sequence, we know that
(s2k) is decreasing. For all k, a2k+3≤ a2k+2, so
s2k+3= s2k+1+ a2k+2− a2k+3≥ s2k+1
Therefore (s2k+1) is increasing.
Next, we show that for all m, n ∈ N, s2m+1≤ s2n. When m = n, we get s2n+1= s2n− a2n+1 ≤ s2n; when
m ≤ n, s2m+1≤ s2n+1≤ s2n; when m ≥ n, s2n≥ s2m≥ s2m+1.
Now, for all k ∈ N, s1≤ s3≤ s5≤ · · · ≤ s2k−1≤ s2k+1≤ s2k ≤ s2k−2≤ s2k−4≤ · · · ≤ s4≤ s2.
(s2k) is decreasing to s1≤ s2k≤ s2 ∀k, so (s2k) is bounded and therefore it converges to some t1. Likewise,
(s2k+1) converges to some t2. Then
t2− t1= lim k→+∞s2k+1− limk→+∞s2k =k→+∞lim (s2k+1− s2k) = lim k→+∞(−1) 2k+1a 2k+1= 0
Therefore t2= t1, and we can show that lim
n→+∞sn= t2= t1.
Corollary 13.1. Let P∞
n=0(−1) na
n be an alternating series. Now, for all k ∈ N, if the sequence (|an|) is
decreasing and lim
n→+∞an= 0, then the series converges.
Example 13.1. Consider an= (1 + 2 · (−1)n) · 2−n; by computing a few terms we see that this series
is alternating. lim
n→+∞an = 0, but |a2k| = 3 · 2
−2k > 2 · 2−2k = 2−(2k−1)= |a 2k−1,
so (|an|) is not decreasing. But an ≤ 3 · 2−n ∀n, which converges because it is a
geometric series, i.e. an converges absolutely by the comparison test.
13.1
Continuity
Definition 13.1. Let S and T be sets. A function (or map) from S to T is a choice of f (s) ∈ T ∀s ∈ S. We denote this by f : S → T .
Lecture 13: Alternating Series, Continuity 56
Definition 13.2. Let S and T be sets and let f : S → T . S is the domain of f (dom F ) and T is the codomain of f .
For us, all functions are of the form f : S → R, where S ⊆ R. If dom(f ) is not specified, it is the largest subset of R where f is well defined.
Example 13.2. f (x) = 1x, dom(f ) = R \ {0}
Example 13.3. g(x) =√4 − x2, dom(g) = [−2, 2]
Definition 13.3. Let f : S → R be a function and let a ∈ S. We say f is continuous at a if there exist sequences (an) such that an ∈ S ∀n and lim
n→+∞an = a. We have
lim
n→+∞f (an) = f (a) = f ( limn→+∞an)
We say f is discontinuous at a if f is not continuous at a. We say f is continuous if f is continuous at x for all x ∈ S.
Theorem 13.2. Let f : S → R and let a ∈ S. Then, f is continuous at a if and only if
∀ > 0, ∃δ > 0 s. t. ∀x ∈ S s. t. |a − x| < δ, |f (a) − f (x)| <
Proof. Suppose the δ − property holds; we want to prove that f is continuous at a.
Let (an) be a sequence in S converging to a. Then, ∀ > 0, ∃δ s. t. ∀x ∈ S with |a − x| < δ, |f (a) − f (x)| < .
Since lim
n→+∞an = a, there exists some N such that |a − an| < δ. Therefore |f (a) − f (an)| < ∀n > N .
Therefore lim
n→+∞f (an) = f (a), so f is continuous at a.
In the reverse direction, we proceed by contraposition, i.e. if the δ − property does not hold, then f is not continuous at a. If the δ − property doesn’t hold, then
∃ > 0 s. t. ∀δ > 0, ∃x ∈ S s. t. |a − x| < δ but |f (a) − f (x)| ≥
For all n ∈ Z>0, pick δ = 1n. Then there exists an∈ S such that |a − an| < δ = 1n, but |f (a) − f (an)| ≥ .
Then (an) is a sequence converging to a, but |f (a)−f (an)| < is never true for any n. So lim
n→+∞f (an) 6= f (a),
Lecture 13: Alternating Series, Continuity 57
Example 13.4. Let f (x) = 2x2+ 1; we want to prove f is continuous. Let a ∈ R. For any sequence (an) converging to a, we have
lim n→+∞f (an) =n→+∞lim 2a 2 n+ 1 = 2 lim n→+∞an 2 + 1 = 2a2+ 1 = f (a) Therefore f is continuous at a.
We can prove the same thing using δ − : work backwards from
|f (a) − f (x)| <
We want to solve for x or |x| or |a − x| in terms of things that do not involve x:
|f (a) − f (x)| = |2a2+ 1 − 2x2− 1| = |2(a + x)(a − x)| = 2|a − x| · |a + x|
We replace |a + x| with something larger. Assume |a − x| < 1. Then
1 > |a − x| ≥ ||a| − |x|| = ||x| − |a|| ≥ |x| − |a|
Therefore
|x| < 1 + |a| |a + x| ≤ |a| + |x| < 2|a| + 1 So
|f (a) − f (x)| < 2|a − x|(2|a| + 1) < |a − x| <
2(2|a| + 1) Now we can begin the proof.
Proof. Let > 0. Define δ = min{1,2(2|a|+1) }. Then ∀x ∈ R such that |a − x| < δ, 2|a − x|(2|a| + 1) < . Also |a − x| < 1, so |x + a| < 2|a| + 1. Therefore
Lecture 13: Alternating Series, Continuity 58
> 2|a − x|(2|a| + 1) > 2|a − x||x + a| = |f (a) − f (x)|
So f is continuous at a by the δ − definition.
Example 13.5. Consider the function
f (x) = (1 xsin 1 x2 x 6= 0 0 x = 0
Show that f is discontinuous at 0.
We want to find some (an) such that lim
n→+∞an = 0, butn→+∞lim f (an) 6= f (0) = 0.
We want to find an (an) such that sin 1/a2n = 1. This gives us
an= 1 pπ 2 + 2πn Then, lim n→+∞an = 0, but f (an) = 1 ansin 1 a2 n = a1 n. Therefore n→+∞lim f (an) = lim n→+∞ 1 an = +∞ 6= f (0). Therefore f is discontinuous at 0. Most familiar functions are continuous everywhere: ex, sin x, cos x, log
bx∀b > 0, xp∀p ∈ R. Note that when
we say a function is continuous everywhere, we mean throughout its domain.
Example 13.6. Consider f (x) = |x|. To show that this is continuous, we want to show that for all (an) converging to a, ||an| − |a|| ≤ |an− a| < (by the triangle inequality). This is
true because f (x) = x is continuous.
13.2
Combining functions
Given f : S → R, g : T → R, 1. (f + g)(x) = f (x) + g(x), with domain S ∩ T 2. (f g)(x) = f (x)g(x), with domain S ∩ T 3. ∀k ∈ R, (kf )(x) = kf (x). 4. fg(x) = f (x)g(x), with domain {x ∈ S ∩ T, g(x) 6= 0}5. max(f, g)(x) = max{f (x), g(x)}; min(f, g)(x) = min{f (x), g(x)}, with domain S ∩ T . 6. (g ◦ f )(x) = g(f (x)), with domain {x ∈ S | f (x) ∈ T }.
Theorem 13.3. Let f : S → R, g : T → R, a ∈ R.
(a) f and g continuous at a implies f + g, f g are continuous at a. (b) ∀k ∈ R, f continuous at a implies kf is continuous at a.
(c) f and g continuous at a and g(a) 6= 0 implies fg is continuous at a. (d) f and g are continuous at a, max(f, g) and min(f, g) are continuous at a. (e) f continuous at a and g continuous at f (a) implies g ◦ f is continuous at a.