Math 115 II 11.1 Sequences, 11.2 , 11.3 The Test, 11.4 Comparison Tests

Mikey Chow

March 16, 2021 Table of Contents Notation for Sequences Using Patterns to Write Formulas for Sequences

Example

Find a formula for the general term an in the sequence below, assuming that the pattern continues. 3 4 5 6 7 8 9 , − , , − , , − , ,... 4 8 16 32 64 128 256 i.e. 3 4 5 6 7 a = , a = − , a = , a = − , a = − ,... 1 4 2 8 3 16 4 32 5 64 If {an} does not have a , then we say the sequence diverges. Example

1 lim = 0 n→∞ n

Limits of Sequences

Definition If {an} is a sequence that gets closer and closer to a number L as n gets large, then we say that the sequence converges and its limit is L and we write lim an = L n→∞ Example

1 lim = 0 n→∞ n

Limits of Sequences

Definition If {an} is a sequence that gets closer and closer to a number L as n gets large, then we say that the sequence converges and its limit is L and we write lim an = L n→∞

If {an} does not have a limit, then we say the sequence diverges. Limits of Sequences

Definition If {an} is a sequence that gets closer and closer to a number L as n gets large, then we say that the sequence converges and its limit is L and we write lim an = L n→∞

If {an} does not have a limit, then we say the sequence diverges. Example

1 lim = 0 n→∞ n Finding Limits of Sequences

Example Find 3n2 + 1 lim n→∞ n2 + n + 1 and √ n lim n→∞ ln n and (−1)n lim n→∞ n Table of Contents Group Problem Solving Question: Suppose whenever I take a bite out of a chocolate bar, I eat exactly half of the remaining chocolate bar. I will not eat the entire chocolate bar after finitely many bites, but will I eat the whole thing after infinitely many bites?

As a group, answer the above question, by finding answers to each of the following subquestions (there is a second set of subquestions on the next slide): A.1. What proportion of the original chocolate bar do I eat on the first bite? A.2. What proportion of the original chocolate bar do I eat on the second bite? A.3. What proportion of the original chocolate bar do I eat on the third bite?

A.4. Let an be the proportion of the original chocolate bar I eat on the nth bite. Give a formula for an. Group Problem Solving Question: Suppose whenever I take a bite out of a chocolate bar, I eat exactly half of the remaining chocolate bar. I will not eat the entire chocolate bar after finitely many bites, but will I eat the whole thing after infinitely many bites?

B.1. What proportion of the original chocolate bar have I eaten in total after the first bite? B.2. What proportion of the original chocolate bar have I eaten in total after the second bite? B.3. What proportion of the original chocolate bar have I eaten in total after the third bite? Pn B.4. Let sn = i=1 ai . What does sn represent? Based on the pattern so far guess a formula for sn.

B.5. What is lim sn? How does this answer the original question? n→∞ ar n−1

Geometric Sequences

A sequence {an} of the form

a, ar, ar 2, ar 3, ar 4,...

i.e. an = , is called a geometric sequence. r is called the geometric ratio. Example Find a and r for the geometric sequence below 3 , −6, 24, −96, 384,... 2 N N X X n−1 2 N−1 ai = ar = a + ar + ar + ··· + ar n=1 n=1

N N X X i−1 2 N−1 N r ai = r ar = ar + ar + ··· + ar + ar n=1 i=n a − ar N

a(1 − r N ) 1 − r

Geometric Sequences

A sequence {an} of the form a, ar, ar 2, ar 3, ar 4,... n−1 i.e. an = ar , is called a geometric sequence. r is called the geometric ratio. The N’th partial sum of the geometric sequence is

sN =

Let’s find a formula for sN by considering rsN :

rsN =

sN − rsN =

sN = ar n−1

Geometric Sequences

A sequence {an} of the form

a, ar, ar 2, ar 3, ar 4,...

i.e. an = , is called a geometric sequence. r is called the geometric ratio. The N’th partial sum of the geometric sequence is a(1 − r N ) s = N 1 − r Example Find the 10’th partial sum for the geometric sequence below 3 , −6, 24, −96, 384,... 2 a 1−r if |r| < 1 DNE if |r| ≥ 1

Geometric Sequences

A sequence {an} of the form

a, ar, ar 2, ar 3, ar 4,...

n−1 i.e. an = ar , is called a geometric sequence. r is called the geometric ratio. The N’th partial sum of the geometric sequence is

N X i−1 2 N−1 sN = ar = a + ar + ar + ··· + ar n=1

a(1 − r N ) s = N 1 − r ( lim sN = N→∞ ∞ X n−1 ar = lim sN N→∞ n=1 |r| < 1 |r| ≥ 1

Geometric Sequences The N’th partial sum of the geometric sequence is

N X a(1 − r N ) s = ar n−1 = a + ar + ar 2 + ··· + ar N−1 = N 1 − r n=1

( a 1−r if |r| < 1 lim sN = N→∞ DNE if |r| ≥ 1

Definition A is an infinite sum

The geometric series converges if and diverges if . Geometric Sequences

Definition A geometric series is an infinite sum

∞ N X n−1 X n−1 ar = lim sN = lim ar N→∞ N→∞ n=1 n=1 The geometric series converges if |r| < 1 and diverges if |r| ≥ 1. Example The geometric series representing the problem we started with today is Geometric Sequences

Definition A geometric series is an infinite sum

∞ X n−1 ar = lim sN N→∞ n=1 The geometric series converges if |r| < 1 and diverges if |r| ≥ 1. Example Write down the geometric series for the geometric sequence below and find it converges or diverges. 3 , −6, 24, −96, 384,... 2 Table of Contents N X a1 + a2 + ··· + aN = an n=1

∞ X an = a1 + a2 + a3 + ··· = lim sN N→∞ n=1

P∞ n=1 an

Partial Sums and Series

Consider a general sequence {an}, i.e.

a1, a2, a3, a4,... The N’th partial sum of the sequence is given by

sN =

Notice that we get a new sequence {sN }.

The series of the sequence is the limit of the sequence sN , which we can think of as the infinite sum

which may or may not exist. If the limit exists, we say the series converges and if the limit does not exist, we say it diverges. Partial Sums and Series The N’th partial sum of the sequence is given by

n X sN = a1 + a2 + ··· + aN = aN n=1

Example

Let an = 2n − 1. Find the first 4 partial sums. Guess a formula for sN based on a pattern. (What does this remind you of?) Example Compute ∞ X 1 1  − n n + 1 n=1 Test

Theorem If a sequence {an} does not converge to 0, then its series diverges, i.e. ∞ X lim an 6= 0 =⇒ an diverges. n→∞ n=1

Remark It is NOT true that ∞ X lim an = 0 =⇒ an converges. n→∞ n=1 Table of Contents Divergence Test

Theorem If a sequence {an} does not converge to 0, then its series diverges, i.e. ∞ X lim an 6= 0 =⇒ an diverges. n→∞ n=1

Example What does the divergence test say about

∞ X 1 1 1 1 = + + + ···? n 1 2 3 n=1 Table of Contents P∞ 1 n=1 n2 Example Determine whether or not ∞ X 1 1 1 1 = + + + ··· n2 12 22 32 n=1 converges. Group Problem Solving Instructions: Following the previous example, determine whether or P∞ 1 not n=1 n converges by doing the following steps on the Jamboard: 1 1. Graph the function f (x) = x and plot the sequence as points on the graph. 2. Draw a Riemann sum that is in fact equal to the series, but is R ∞ 1 an overestimate of the integral 1 x dx. P∞ 1 3. Using Step 2., determine the convergence of n=1 n . Do a similar procedure to the above to determine the convergence P∞ √1 of n=1 n . As a group, guess what the general rule is:

∞ ( X 1 converges for = np n=1 diverges for ∞ X Z ∞ an converges if and only if f (x)dx converges. n=1 1

The Integral Test

Theorem (The Integral Test)

Suppose the sequence an is given by f (n) where f is a continuous, positive, decreasing sequence on [1, ∞). Then

Example Determine the convergence of the series of the sequence e−1, 2e−2, 3e−3,... Table of Contents ∞ ∞ X X 0 ≤ an ≤ bn n=1 n=1

P∞ P∞ n=1 an n=1 bn P∞ P∞ n=1 bn n=1 an

The (Basic) Comparison Test

Theorem (The (Basic) Comaprison Test)

Suppose {an} and {bn} are two sequences such that 0 ≤ an ≤ bn for all n. Then

In particular, If converges, then converges also. If diverges, then diverges also. (What does this remind you of?) a lim n = L n→∞ bn where L is a finite number and L 6= 0.

∞ ∞ X X an converges if and only if bn converges. n=1 n=1

The Limit Comparison Test

Theorem (The Limit Comaprison Test)

Suppose {an} and {bn} are two sequences such that

Then Group Problem Solving Instructions: For each pair of series below, determine the convergence of one of them without using a comparison test. Then use a comparison test with these two series to determine the convergence of the other.

∞ ∞ X 1 X 1 2n 2n + 1 n=1 n=1 ∞ ∞ X 1 X 1 2n 2n − 1 n=1 n=1 ∞ ∞ X 1 X n2 + 1 n2 n4 − n2 + 1 n=1 n=1 ∞ ∞ √ X 1 X i 2 + 1(i 3 + 3) n i 5 − i 4 + i 3 n=1 i=1 Group Problem Solving Instructions: For each pair of series below, determine the convergence of one of them without using a comparison test. Then use a comparison test with these two series to determine the convergence of the other.

∞ ∞ X sin n + 2 X 1 n2 n2 n=1 n=1 ∞ ∞ X n sin2 n + cos n + 3 X 1 n4 + 5 − 4 cos n n3 n=1 n=1 ∞ ∞ X X 12e−n e−n log(n + 1) n=1 n=1 ∞ ∞ X e10−n X 1 n5 N5 n=1 N=1