Sequences And Series Review

Sequences and Series Review Two very important (distinct) mathematical objects are sequences and series. A sequence is P an infinite list of numbers, whereas a series is an infinite ∞ sum. With a series n=1 an , though, we can associate Pn two sequences: 1) its sequence of partial sums (Sn ), where Sn = k=1 ak 2) its sequence of terms (an ). Our primary concern with both sequences and series is convergence. A sequence (an ) converges Pnif limn→∞ an = L for some real number L. A series converges if limn→∞ k=1 ak = L for some real number L; this is the same as saying the sequence of partial sums (Sn ) converges. If a sequence or series does not converge, we say it diverges. We have developed many tools for determining whether a given sequence or series converges. The ultimate goal of all these tests is applying them to Taylor series. We know that if the Taylor series of f at a converges for a given value of x, then it converges to f (x). This is why we only test for convergence rather than try to evaluate these series: once we have convergence, we know right away what it converges to. Sequence list

Series sum

Written

(an ) a1 , a2 , a3 , . . .

P∞

Converges if

limn→∞ an = L

Pn limn→∞ k=1 ak = L limn→∞ Sn = L

Tests

Limit Laws, Substitution Law Squeeze Law Limits of Functions and Sequences Bounded and Monotone L’Hˆopital’s Rule

nth Term Test Integral Test Comparison Test Limit Comparison Test Alternating Series Test Ratio Test Root Test p-Series Geometric Series

What it is

n=1 an a1 + a2 + a3 + · · ·

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Sequences

Limit Laws, Substitution Law These laws tell us how to compute limits of sequences. If we can determine what a sequence converges to, then trivially it converges. Recall the Substitution Law says if 1) limn→∞ an = L 2) f is continuous at L, then limn→∞ f (an ) = f (L). Squeeze Law The Squeeze Law is very intuitive: if an ≤ bn ≤ cn and limn→∞ an = limn→∞ cn = L, then limn→∞ bn = L also. Often we apply this with sin and cos, bounding them between −1 and 1. Limits of Functions and Sequences This is the idea that if limx→∞ f (x) = L, then limn→∞ f (n) = L also. For example, limx→∞ 2 + 1/x = 2, so limn→∞ 2 + 1/n = 2. This rule essentially exists to justify using L’Hˆopital’s Rule on sequences. Bounded and Monotone By a theorem, any sequence (an ) which is bounded (for some positive number M , and for all n, −M ≤ an ≤ M ) and monotone (for all large enough n, either the an ’s are increasing or decreasing) converges. This theorem can only tell us that a sequence converges, not to what it converges. L’Hˆopital’s Rule L’Hˆopital’s Rule says that if limn→∞ f (n) and limn→∞ g(n) are both 0 or both ∞, then f ′ (n) f (n) = lim ′ . lim n→∞ g (n) n→∞ g(n) Using standard techniques, we can use L’Hˆopital’s Rule to determine limits of indeterminate forms 0 · ∞, 0∞ , 1∞ , and ∞0 .

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Test Limit Laws

Sequence (an )

Check lim an = L

Limit L

Substitution Law

(f (an ))

lim an = L f continuous at L

f (L)

Squeeze Law

(bn )

an ≤ bn ≤ cn limn→∞ an = limn→∞ cn = L

L L

Functions and Sequences

(f (n))

limx→∞ f (x) = L

Bounded and Monotone

(an )

−M ≤ an ≤ M for all n an monotone

L’Hˆopital’s Rule

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(f (n)/g(n))

lim f (n) = lim g(n) = 0 or lim f (n) = lim g(n) = ∞

No info lim f ′ (n)/g ′ (n)

Series

nth Term Test If

P∞

n=1

an converges, to L say, then

lim an = lim Sn − Sn−1 = lim Sn − lim Sn−1 = L − L = 0.

n→∞

n→∞

n→∞

n→∞

The nth Term Test is the P contrapositive of this statement: if limn→∞ an does ∞ not exist or is not 0, then n=1 an diverges. This test is easy to apply, and should probably be the first test applied to determine if a certain series converges or diverges. Integral Test P∞ If n=1 an is a series and f (x) is a function such that (1) f (x) is positive, decreasing, and continuous (2) f (n) n, R∞ P∞= an for all then n=1 an and 1 f (x) dx both converge or both diverge. Comparison Test P∞ P∞ The Comparison Test says that if n=1 an and n=1 bn are positive-term series, P∞ P∞ (1) if an ≤ bn for all n and Pn=1 bn converges, thenP n=1 an converges ∞ ∞ (2) if an ≥ bn for all n and n=1 bn diverges, then n=1 bn diverges.

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P∞ How you use the Comparison Test depends on whether you think n=1 an converges or diverges. If you think it converges, try to find a convergent series P∞ b with an ≤ bn , and if you think it diverges, find a divergent series n n=1 P∞ n=1 bn with an ≥ bn . Limit Comparison Test P∞ P∞ If n=1 an and n=1 bn are positive-term series an /bn = L P∞ P∞and limn→∞ with L a positive real number (note L 6= 0), then n=1 an and n=1 bn both converge or both diverge. Alternating Series Test P∞ If n=1 an is an alternating series (the terms an alternate sign) and (1) limn→∞ an = 0 (2) |aP n | is decreasing ∞ then n=1 an converges. P∞ The alternating series test comes with a remainder estimate. If n=1 an is an alternating series converging to L, then |L −

N X

n=1

an | ≤ |aN +1 |.

Using this remainder estimate, it is easy to approximate the value of an alternating series to any degree of accuracy: if you want to know the value to within an error of E, find an N such that |aN +1 | ≤ E, and then simply sum the first N terms. Geometric Series P∞ A geometric series is a series n=1 an such that an = a1 rn−1 for some real number r. Not only do we know when these converge, but we also have a formula for their value:P ∞ (1) if |r| < 1, Pn=1 an = a1 /(1 − r) ∞ (2) if |r| ≥ 1, n=1 an diverges. Ratio Test The RatioPTest says that if ρ = limn→∞ |an+1 /an |, then ∞ (1) if ρ < 1, Pn=1 an converges absolutely ∞ (2) if ρ > 1, n=1 an diverges. The test is inconclusive if ρ = 1. P∞ The Ratio Test allows us to say n=1 an is almost a geometric series with ratio ρ, since for a geometric series ρ = |r|. This is why the convergence criteria for Ratio Test and geometric series are similar.

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The Ratio Test is particularly suited for series involving factorials and powers, since most of it will cancel when computing the limit. Root Test p The Root Test says if ρ = limn→∞ n |an |, then the conclusions of the Ratio Test hold. The Ratio Test and the Root Test always give the same value of ρ. √ When using the Root Test, one often has to compute a limit such as limn→∞ n n. The secret for these sort of limits is to convert them to powers of e: √ n n = n1/n = (eln n )1/n = eln n/n . √ Since limn→∞ ln n/n = 0, the Substitution Rule says lim n n = e0 = 1. p-Series P∞ 1 The series n=1 np converges if p > 1 and diverges if p ≤ 1. This is a consequence of the Integral Test.

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Test

Check

Conclude

nth Term

limn→∞ an 6= 0

Diverges

Integral

f (x) > 0, cont., decr. f (n) = an R∞ if R1 f (x)dx converges ∞ if 1 f (x)dx diverges

Comparison

Limit Comparison

a n ≤ bn P ∞ n=1 bn converges a n ≥ bn P ∞ n=1 bn diverges

Converges Diverges

limn→∞ an /bn = L, L 6= 0 P∞ bn converges Pn=1 ∞ n=1 bn diverges

Alternating Series

Converges Diverges

|an | decreases limn→∞ an = 0

Converges Diverges

Converges

Geometric

|r| < 1 |r| ≥ 1

Converges to a1 /(1 − r) Diverges

Ratio

ρ<1 ρ>1

Converges (abs.) Diverges

Root

ρ<1 ρ>1

Converges (abs.) Diverges

p-Series

p>1 p≤1

Converges Diverges

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Taylor series

The Taylor series for a function f (x) at a point a is simply ∞ X f (k) (a)

k=0

k!

(x − a)k .

The Maclaurin series is just the Taylor series with a = 0. An important feature of the Taylor series is that if it converges for a given value of x, then it converges to f (x).

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To compute the Taylor series for f (x) at a, we just have to find the derivatives of f , evaluate them at a, and plug into the formula. Taylor series have nice convergence properties. For example, they always converge on an interval (the interval of convergence) whose midpoint is a. The radius of convergence is half the length of the interval of convergence. Related to Taylor series are Taylor polynomials. The nth degree Taylor polynomial of f (x) at a is Pn (x) =

n X f ( k)

k!

k=0

(x − a)k ,

so Pn (x) is just the first n + 1 terms of the Taylor series. Taylor’s Theorem tells us that, as long as we can differentiate f (x) n + 1 times, f (x) − Pn (x) = Rn (x), where f (n+1) (z) (x − a)n+1 Rn (x) = (n + 1)! for some z between x and a. In general, it is impossible to solve for z here. In practice, we try to bound Rn (x) by treating it as a function of z and finding where it achieves its extrema.

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Convergence

P∞ There arePtwo types of convergence for a P series n=1 an : absolute and con∞ ∞ ditional. n=1 |an | converges. All absolutely n=1 an converges absolutely if P∞ convergent sequences converge. Conditional convergence happens when n=1 |an | P∞ diverges but n=1 an converges. P ∞ harmonic series n=1 (−1)n /n converges conditionally, whereas P∞The alternating n n=1 (−2/3) converges absolutely.

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Problems

1. Determine the limits√of the following sequences: (1) an = ln n/n n n an = (4/3)n (3) an = nln n (4) an = (2/3)n n2 (5) an = 1,000,000 n! sin n (7) an = √1n . an = ln ln ln(n)

(2) (6)

p √ 2. Show that the sequence 25 − 1/ n converges by first applying the bounded and monotone test, then determine its limit. 3.

Does the series

P∞

n n=1 (−1/2)

+ 1/n converge?

P∞ 1 4. Does n=1 n ln n ln ln n converge? Determine all values of p for P∞the series1 which n=1 n ln n(ln ln n)p converges. 7

5. Determine if the following series converge or diverge. (1) ∞ X 1 n=1

ln n

(2)  n ∞ X 1 2 n 3 n=1 (3)

√ ∞ X | sin(n4 ln(cos2 ( n)) + 1)| n17/16 n=1

(4) ∞ X ln n √ n n=1

(5) ∞ X ln n √ n n n=1

(6) ∞ X 3n n1000 10n n=1

(7) ∞ X

n=1

(2/3)n − (3/5)n .

6. Use the Limit Comparison Test to determine if the following series converge or diverge. (1) √ ∞ X 4 n + 78 13n2 + 42 n=1 (2) ∞ X n+1 n2 n=1

(3) ∞ X 4n + 16n/2 7n n=1

(4)

P∞

n=1

1 − 1/n.

7. P Determine if √ the following series converge or diverge. ∞ (1) n=1 (−1)n / n 8

P∞ n (2) n=1 (−1) − 1/n P∞ p n (3) n=1 (−1)n n2 . 8. Determine the value of the alternating harmonic series within .2.

P∞

n n=1 (−1) /n

to

9. Find the Taylor series of ln(1 + x) at a = 0 and determine its interval of convergence. 10.

Show that the Taylor series for ex , sin(x), and cos(x) converge for all real x.

11. (1)

Determine the interval and radius of convergence of the following series ∞ X (x − 3)n 2n n=1

(2) ∞ X (x + 1)n n! n=1

(3) ∞ X xn n n=1

(4) ∞ X (−2)n + 3n n x . 7n n=1

12. Use your answer from Problem 9 to find ln(3/2) accurate to 3 decimal places. 13. Use L’Hˆopital’s Rule to find limx→0 (x − sin(x))/x3 . Then find the Taylor series of (x − sin(x))/x3 . What is its limit as x goes to 0? 14. Find the fourth degree Taylor polynomial of sin(x) at a = 0 and also the remainder estimate. How many decimal places of accuracy can you guarantee if you approximate sin(1/2) with P4 (1/2)? 15. Suppose the Maclaurin series for f (x) has interval of convergence (−2, 2]. What is the radius of convergence of the Maclaurin series of f (3x)? 16. Without calculating derivatives, find the Taylor series of the following functions at a = 0. 2 (1) e5x (2) sin(x9 ) (3) x/(1 − x)2 (4) (sin(x) − x)/x2 . 17.

Show that

R

sin(x)dx = − cos(x) + C using Taylor series. 9