Geometric Series and the Ratio and Root...

MATH 4540: Analysis Two

Geometric Series and the Ratio and Root Test

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

February 1, 2019


Outline

1 Geometric Series

2 The Ratio Test

3 The Root Test

4 Examples

5 A Taste of Power Series

6 Homework


Geometric Series

Theorem

Geometric Series:∑∞n=0 r

n converges if −1 < r < 1 to 1/(1− r).∑∞n=0 r

n diverges if r 6∈ (−1, 1).

Proof

(1):We show

∑∞n=0 r

n converges if −1 < r < 1 to 1/(1− r). We alreadyknow the partial sum Sn = 1 + r + . . .+ rn = (1− rn+1)/(1− r) form aprevious induction argument. We also know that rn+1 → 0 as n→∞since −1 < r < 1. Thus,

∑∞n=0 r

n converges to 1/(1− r).

(2):If r = 1, we have the series 1 + 1 + · · ·+ 1 + . . . which has partial sumsSn = n. The partial sums are unbounded and so the series does notconverge.

If r = −1, we have seen already that∑∞

n=0(−1)n diverges by oscillation.


Geometric Series

Proof

If r > 1, we have Sn = 1 + r + . . .+ rn = (rn+1 − 1)/(r − 1) and we seelimn→∞ Sn =∞ since rn+1 →∞ and r − 1 > 0. So this series diverges.

If r < −1, the limn→ rn does not exist and so by the nth term test, theseries must diverge.

Comment: note if −1 < r < 1(1)

∑∞n=1 r

n = 1/(1− r)− 1(2)

∑∞n=2 r

n = 1/(1− r)− 1− r(3)

∑∞n=3 r

n = 1/(1− r)− 1− r − r2

So(1)

∑∞n=1(1/3)n = 1/(1− (1/3))− 1 = 3/2− 1 = 1/2

(2)∑∞

n=2(1/3)n = 1/(1− (1/3))− 1− (1/3) = 1/2− 1/3 = 1/6

(3)∑∞

n=3 rn = 1/(1− (1/3))− 1− (1/3)− (1/9) = 1/6− 1/9 = 1/(18)


The Ratio Test

Theorem

The Ratio Test:Let

∑∞n=1 an be a series with positive terms. and assume

limn→∞ an+1/an = ρ.(1) if 0 ≤ ρ < 1, the series converges.(2) if ρ > 1, the series diverges.(3) if ρ = 1, we don’t know.

Proof

(1):We assume 0 ≤ ρ < 1. Then, we see ρ < (1 + ρ)/2 < 1. Let the value(1 + ρ)/2 be the “r” in the geometric series

∑∞n=1 r

n which convergessince 0 < r < 1. We will use the comparision test to show our seriesconverges.

Let ε = (1− ρ)/2 which is a nice positive number. Then,


The Ratio Test

Proof

ε+ ρ = (1 + ρ)/2 = r < 1

Since limn→∞ an+1/an = ρ, for this ε, there is an N so that

n > N =⇒∣∣∣∣an+1

an− ρ∣∣∣∣ < ε =

1− ρ2

Thus, an+1/an < ε+ ρ = (1 + ρ)/2 = r < 1 for all n > N. We see

aN+2 < r aN+1

aN+3 < r aN+2 < r (r aN+1) = r2aN+1

aN+4 < r aN+3 < r (r2 aN+1) = r3aN+1

...

aN+` < r `−1aN+1 =aN+1

rN+1r `+N

for ` > 1.


The Ratio Test

Proof

Now let k = N + ` and c = aN+1/rN+1. We have shown therefore that

ak < crk for k > N + 1.

It is easy to see the series c∑∞

k=N+2 rn converges by the limitcomparison test, so by the comparison test, we have shown

∑∞k=N+2 ak

converges which also tells us∑∞

k=1 ak converges.(2):If ρ > 1, choose ε = (ρ− 1)/2 > 0 as ρ > 1. Then, there is an N so that

n > N =⇒∣∣∣∣an+1

an− ρ∣∣∣∣ < ε =

ρ− 1

2

This implies for n > N that

1 + ρ

2= ρ− ε <

an+1

an< ρ+ ε

Thus, letting r = (1 + ρ)/2 > 1, we have


The Ratio Test

Proof

aN+2 > r aN+1

aN+3 > r aN+2 > r (r aN+1) = r2aN+1

aN+4 > r aN+3 > r (r2 aN+1) = r3aN+1

...

aN+` > r `−1aN+1 =aN+1

rN+1r `+N

for ` > 1. Let C = aN+1/rN+1 and k = N + `. Then, we have ak > C rk

for all k > N + 1.

Now apply the comparison test. Since∑∞

k=N+2 rk diverges since r > 1

and we know C∑∞

k=N+2 rk diverges, by the comparison test∑∞

k=N+2 akdiverges.

(3):If ρ = 1, the series could converge or diverge.


The Ratio Test

Proof

(i) the series∑∞

n=1 1/n2 converges but ρ = 1.(ii) the series

∑∞n=1 1/n diverges but ρ = 1.

So if ρ = 1, we just don’t know.

We can say more:

Theorem

The Ratio Test for Absolute Convergence:Let

∑∞n=1 an be a series with nonzero terms. Assume limn→∞ | an+1

an|

converges to the value ρ.(1) if 0 ≤ ρ < 1, the series converges.(2) if ρ > 1, the series diverges.(3) if ρ = 1, we don’t know.


The Ratio Test

Proof

(1):If 0 ≤ ρ < 1, we can apply the previous theorem to see

∑∞n=1 |an|

converges. But if a series converges absolutely, it also converges.(2):If ρ > 1, the arguments in the proof above show us there is an N so wecan write |ak | ≥ Crk for k > N + 1 for r = (1 + ρ)/2 > 1. Now if∑∞

k=N+2 ak converged, this implies ak → 0. But if that were true,

|ak | → 0 too. But the inequality |ak | ≥ Crk with r > 1 says the sequence(|ak |) can not converge. Thus,

∑∞k=N+2 ak must also diverge. Of course,

this tells us∑∞

n=1 an also diverges.(3):The same examples as before show us why we can’t conclude anythingabout convergence or divergence of the series if ρ = 1.


The Root Test

Theorem

The Root Test:Let

∑∞n=1 an be a series with positive terms. Assume limn→∞(an)1/n

exists and equals ρ.(1) if 0 ≤ ρ < 1, the series converges.(2) if ρ > 1, the series diverges.(3) if ρ = 1, we don’t know.

Proof

(1):We assume 0 ≤ ρ < 1. Choose ε = (1− ρ)/2 > 0. Then there is an N sothat

n > N =⇒ |(an)1/n − ρ| < ε = (1− ρ)/2

We can rewrite this: for n > N

−(1− ρ)/2 < (an)1/n − ρ < (1− ρ)/2 =⇒ (an)1/n < (1 + ρ)/2 < 1


The Root Test

Proof

Let r = (1 + ρ)/2 < 1.

(aN+1)1/(N+1) < r =⇒ aN+1 < rN+1

(aN+2)1/(N+2) < r =⇒ aN+2 < rN+2

...

(aN+`)1/(N+`) < r =⇒ aN+` < rN+`

for ` ≥ 1. Letting k = N + `, we have ak < rk for k > N. Since0 < r < 1, the series

∑∞k=N+1 r

k converges and so by comparison, sodoes

∑∞k=N+1 ak . This shows

∑∞k=1 ak converges.

(2):Now ρ > 1. This argument is very similar to the one use for the proof ofthe ratio test.


The Root Test

Proof

Let ε = (ρ− 1)/2 > 0. Then there is an N so that

n > N =⇒ |(an)1/n − ρ| < ε = (ρ− 1)/2

We can rewrite this as

(1− ρ)/2 < (an)1/n − ρ < (ρ− 1)/2 =⇒ (an)1/n > (1 + ρ)/2 > 1

for n > N. Let r = (1 + ρ)/2.

Then, we have the inequality an > rn for all n > N.

Since∑∞

n=N+1 rn diverges since r > 1, by the comparison test, we know∑∞

n=N+1 an diverges and so the original series diverges too.


The Root Test

Proof

(3):If ρ = 1, we have limn→∞ n1/n = 1. Look at the series

∑∞n=1 1/n. We

know this diverges and ρ = 1.

Now look at the series∑∞

n=1 1/n2. This converges andlimn→∞(n2)1/n = limn→∞ n2/n. But limn→∞ n2/n = limn→∞(n1/n)2.Since the function f (x) = x2 is continuous, we also knowlimn→∞(n1/n)2 = ( limn→∞ n1/n )2 = 1.

So we just don’t know whether the series converges or diverges if ρ = 1.

We can restate this result for series whose terms are not necessarily

non-negative.


The Root Test

Theorem

The Root Test for Absolute Convergence:Let

∑∞n=1 an be a series. Assume limn→∞ |an|1/n exists and equals ρ.

(1) if 0 ≤ ρ < 1, the series converges.(2) if ρ > 1, the series diverges.(3) if ρ = 1, we don’t know.

Proof

(1):If 0 ≤ ρ < 1, the previous theorem tells us

∑∞n=1 |an| converges

absolutely implying convergence.(2):If ρ > 1, the earlier arguments imply is an N so |ak | ≥ rk for k > N forr = (1 + ρ)/2 > 1. If

∑∞k=N+1 ak converged, then ak → 0 and |ak | → 0

too. But |ak | ≥ rk with r > 1 says the sequence (|ak |) can not converge.Thus,

∑∞k=N+1 ak must also diverge and so the original series diverges

too.


The Root Test

Proof

(3):The same examples as before show us why we can’t conclude anythingabout convergence or divergence of the series if ρ = 1.

Let’s summarize what we know:(1) We can check convergence with comparison tests. If the terms arenot non-negative, we can test for absolute convergence which will implyconvergence.

(2) The ratio and root test are also good tools. We can check for

absolute convergence if the terms are not non-negative to see if the series

converge. But the ratio test requires that none of the terms are zero.


Examples

Example

What is the sum of

(3/4)5 + (3/4)6 + . . .+ (3/4)n + . . .?

This is a geometric series with r = 3/4 missing the first 5 terms. So thesum of the series is

1/(1− (3/4))− 1− (3/4)− (3/4)2 − (3/4)3 − (3/4)4.

Note the five terms we subtract could be written

1 + (3/4) + (3/4)2 + (3/4)3 + (3/4)4 = (1− (3/4)5)/(1− (3/4))

but this is not necessarily an improvement!


Examples

Example

(1):What is the sum of

∑∞k=1 (1/(3k)− 1/(4k))?

To do this we have to write the inner part as a single fraction. We find

∞∑k=1

( 1/(3k)− 1/(4k) ) =∞∑k=1

k/(12k2) =∞∑k=1

1/(12k)

which diverges as it is multiple of the harmonic series.(2):Does

∑∞k=5( (2k3 + 3)/(k5 − 4k4) ) converge?

This looks like the series∑

1/k2 so a good choice would be the limitcomparison test. But that gives

limk→∞

(2k3 + 3)/(k5 − 4k4) /(1/k2 = limk→∞

(2k5 + 3k2)/(k5 − 4k4) = 2

Thus, since the limit is positive and finite, our series converges.


Examples

Example

Note the ratio test will fail. We find

limk→∞

(2(k + 1)3 + 3)/((k + 1)5 − 4(k + 1)4)

(2k3 + 3)/(k5 − 4k4)

= limk→∞

(2(k + 1)3 + 3) (k5 − 4k4)

((k + 1)5 − 4(k + 1)4) (2k3 + 3)

= limk→∞

2k8 + lower power terms of k

2k8 + lower power terms of k= 1

and so the ratio test fails. Of course, it would be insane to apply the root

test here!


Examples

Example

Does∑∞

k=1 1/( (2k + 1)2k ) converge?

We’ll try the ratio test here:

limk→∞

1/( (2(k + 1) + 1)2k+1 )

1/( (2k + 1)2k )= lim

k→∞

(2k + 1)2k

(2k + 3)2k+1= lim

k→∞

2k + 1

(2k + 3)2

The limit is 1/2 and hence, this series converges by the ratio test.


A Taste of Power Series

A series of the form∑∞

n=0 anxn is called a power series centered at

x = 0. Note no matter what the numbers an are, this series alwaysconverges at x = 0.

Let’s assume the terms an are not zero and thatlimn→∞ |an+1/an| = ρ > 0. We can check for the convergence anddivergence of this series using the ratio test.

limn→∞

∣∣∣∣an+1xn+1

anxn

∣∣∣∣ = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ |x | = ρ|x |

By the ratio test, the series converges when ρ|x | < 1 and diverges whenρ|x | > 1. So we have(1) The series converges when −1/ρ < x < 1/ρ(2) The series diverges when x > 1/ρ or x < −1/ρ(3) We don’t know what happens at x = ±1/ρ.

The number 1/ρ is called R, the radius of convergence of the power

series. Hence, this power series defines a function f (x) locally at x = 0

with radius of convergence R > 0. Any function f (x) which has a power

series which matches it locally at x = 0 is said to be analytic at x = 0.



The Derived series obtained from f (x) =∑∞

n=0 anxn is the series∑∞

n=0 n anxn−1. We are still assuming limn→∞ |an+1/an| = ρ > 0, so we

can check the convergence of this series using the ratio test too.

limn→∞

∣∣∣∣ (n + 1) an+1xn

n anxn−1

∣∣∣∣ = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ limn→∞

(n + 1

n

)|x | = ρ|x |

Hence, this series converges when |x | < R = 1/ρ also.

We see the derived series defines a function g(x) on the circle (−R,R).An obvious question is this: Is f ′(x) = g(x) on (−R,R)? That is does

f ′(x) =

( ∞∑n=0

anxn

)′=

∞∑n=0

(anx

n

)′=∞∑n=0

n anxn−1 = g(x)?

Note this is an interchange of limit order idea. Is it possible tointerchange the limit called differentiation with the limit called series?When we can do this, this is called differentiating the series term byterm.



We will find that we can differentiate and integrate a power series termby term as many times as we like and the radius of convergence remainsthe same! These are the kinds of series we construct using Taylorpolynomials and the resulting series is called a Taylor Series expansionof f (x) at x = 0.

We have a long way to go to be able to prove this sort of theorem. Butthis is the kind of manipulation we must do to use power series to solveODEs with non constant coefficients!

Another kind of series that is very useful isa0 +

∑∞n=1( an cos(nπx/L) + bn sin(nπx/L) ) which is called a Fourier

series. Note its building blocks are sin and cos functions rather thanpowers of x .


Homework

Homework 6

6.1 Determine if∑∞

n=1 n!/2n converges.


n=1 ( (−1)n(2n + 5) )/n! converges.


n=1 2/(n3 + 5n − 2) converges.

6.4 Determine for what values of x the series∑∞n=0 (−1)nx2n/( (2n)! ) converges and for what values of x

it diverges.

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