Convergence Tests We are now interested in developing tests to tell - - PowerPoint PPT Presentation

Convergence Tests

We are now interested in developing tests to tell whether or not a series converges, without having to guess at its sum. The first such test is entirely a negative result.

• Theorem. (the divergence test).

(a) If , then the series diverges. (b) If , then the series may either converge

• r diverge.

lim uk k ≠ →∞ uk ∑ lim uk k = →∞ uk ∑

The divergence test is strictly a negative test. It can be used to show that a series diverges, but it cannot be used to show that a series converges.

• Proof. (a) It is clear by definition that uk = sk − sk-1. If the series

converges to s, then we have lim lim . 1 s s s k k k k = = − →∞ →∞ Thus lim lim ( ) lim lim 0. 1 1 u s s s s s s k k k k k k k k k = − = − = − = − − →∞ →∞ →∞ →∞ (b) Follows by showing both a convergent series and a divergent series for which lim 0. uk k = →∞ The two series are 1 1 1 1 2 4 2k + + + + +

• and

1 1 1 1 2 3 k + + + + +

Problem. (a) Show that the series diverges. 2 2 4 6 2 2 3 4 5 2 1 k k k k k ∞ = + + + + + ∑ + + =

• 2

2 lim lim 2 0 2 2 1 k k k k k = = ≠ + →∞ →∞ +

• Solution. The nth term sequence has the limit

Therefore the series diverges.

Problem. (a) Show that the series diverges. 1 3 5 9 1 1 1 2 4 8 2 2 1 k k k ∞   + = + + + + + + ∑     =

• 1

lim 1 1 0 2k k + = ≠ →∞

• Solution. The nth term sequence has the limit

Therefore the series diverges.

Theorem (a) If Σuk and Σvk are convergent series, then Σ(uk + vk) and Σ(uk − vk) are also convergent, and we have (b) If c is a nonzero constant, then the series Σuk and Σcuk both converge or both diverge. In the case of convergence we have ( ) 1 1 1 u v u v k k k k k k k ∞ ∞ ∞ + = + ∑ ∑ ∑ = = = ( ) 1 1 1 u v u v k k k k k k k ∞ ∞ ∞ − = − ∑ ∑ ∑ = = = 1 1 cu c u k k k k ∞ ∞ = ∑ ∑ = =

• Example. Find the sum of the series

5 7 1 3 4 1 k k k ∞ − ∑ − = The series breaks down into the difference of two series, 5 7 and . 1 3 4 1 1 k k k k ∞ ∞ ∑ ∑ − = = The first series 5 5 5 5 5 3 9 27 3 3 1 k k k ∞ = + + + + ∑ =

• is geometric with a = 5/3 and r = 1/3. It therefore converges to

5 5 3 . 1 2 1 3 = − The second series 7 7 7 7 1 4 16 4 1 k k ∞ = + + + ∑ − = is also geometric with a = 7 and r = 1/4. It therefore converges to

7 7 28. 1 3 3 1 4 4 = = − Thus by (a) the original series converges to 5/2 − 28/3 = (15 − 56)/6 = −41/6. Solution. 7 1 7 . 1 1k k k k ∞   ∞ = ∑   ∑   = =   We showed before that the series 1 1k k ∞   ∑     =  

• diverges. This series is a constant multiple of a

divergent series and so diverges by part (b).

• Example. Determine whether the following series converges or

diverges. 7 7 7 7 7 2 3 4 1k k ∞ = + + + + ∑ =

Theorem Convergence or divergence is not affected by deleting a finite number of terms from a series; in particular, for any positive K, the series and both converge or both diverge. 1 2 3 1 u u u u k k ∞ = + + + ∑ =

• 1

2 u u u u k K K K k K ∞ = + + + ∑ + + =

• Warning. The above theorem says that convergence is

unaffected by removal of a finite number of terms. However, the sum usually is changed.

• Example. Determine whether the following series converge or

diverge. (a) (b) 1 1 1 1 8 9 10 8k k ∞ = + + + ∑ =

• Solution. (a)

is just the divergent series 1 1k k ∞   ∑     =   With the first 7 terms missing. By the theorem above 1 1 1 1 8 9 10 8k k ∞ = + + + ∑ =

• above, it must also diverge.

1 1 1 4 3 5 7 2 4 8 + − + + + + +

• Example. Determine whether the following series converge or

diverge. (a) (b) 1 1 1 1 8 9 10 8k k ∞ = + + + ∑ =

• 1 1 1

2 4 8 + + +

• Solution. (b) The series

is a geometric series with a = 1/2 and r = 1/2. It therefore converges. Thus by the previous theorem the sequence 1 1 1 4 3 5 7 2 4 8 + − + + + + + 1 1 1 4 3 5 7 2 4 8 + − + + + + + must also converge. Of course the sums are different. The geometric series formed when the first 4 terms are dropped has sum (1/2)/(1 − 1/2) = 1, and therefore the original series converges to 9 + 1 = 10.

The integral test

Let us compare the series with the improper integral 1 2 1k k ∞ ∑ = 1 . 2 1 dx x ∞ ∫ Clearly, if the series converges, the sum of the series is greater than the integral, while if the integral diverges, so does the series. 1 1/4 1/9

On the other hand, the diagram below shows that if the integral converges, so does the series and therefore so does the

• riginal series. On the other hand, if the series diverges, then so

must the integral. 1 2 2k k ∞ ∑ = 1/4 1/9 1/16

Theorem (The Integral Test) Suppose that a series Σuk = Σf(k) has positive terms, and that f(x) is the formula resulting from replacing k by x in the expression for the terms of the series. If the function with formula f(x) is decreasing and continuous on the interval [a, ∞), then the series and the integral both converge or both diverge. This leads to the following theorem. 1 uk k ∞ ∑ = ( ) f x dx a ∞ ∫ Note that the value of a does not matter.

• Example. Use the integral test to determine whether the

following series converge or diverge. 1 (a) 1k k ∞ ∑ = 1 (b) 2 1k k ∞ ∑ = 1 (c) , a positive number. 1 p p k k ∞ ∑ = Solution (a). lim lim ln( ) . 1 1 r dx dx r x x r r ∞ = = =∞ ∫ ∫ →∞ →∞ Since the improper integral converges, so does the series.

• Example. Use the integral test to determine whether the

following series converge or diverge. 1 (a) 1k k ∞ ∑ = 1 (b) 2 1k k ∞ ∑ = Solution (b). 1 1 lim lim lim 1 1. 2 2 1 1 1 r r dx dx x r r r r x x ∞     = = − = − = ∫ ∫     →∞ →∞ →∞     Since the improper integral converges, so does the series.

• Note. The series does not converge to 1. The theorem

does not say that the series and the improper integral have the same value when they converge. 1 (c) , a positive number. 1 p p k k ∞ ∑ =

• Example. Use the integral test to determine whether the

following series converge or diverge. 1 (a) 1k k ∞ ∑ = 1 (b) 2 1k k ∞ ∑ = Solution (c). We already know that this diverges if p = 1. If p is not 1, then we have 1 1 1 lim lim lim 1 1 1 1 1 1 1 if 1 1 if 1 p r r dx x p p x dx r p p p r r r x p p p −   ∞ − −     = = = − ∫ ∫   − −   →∞ →∞ →∞      >  − =  ∞ <  1 (c) , a positive number. 1 p p k k ∞ ∑ =

Theorem (The p - series test) converges if p > 1 and diverges if 0 < p ≤ 1. Thus we have the following result of using the integral test

• n series .It is called the p - series test.

1 1 p k k ∞ ∑ = 1 1 1 1 1 2 3 1 p p p p k k k ∞ = + + + + + ∑ =

• Problem. Use any method to determine if the following series

converges or diverges. 3 5 1 k k ∞ ∑ =

• Solution. The first thing we usually do is check the divergence
• test. Clearly the terms in this series do converge to 0, so the

divergence test gives us no information. We can write the series as 3 1 3 5 5 1 1 k k k k ∞   ∞ = ∑   ∑   = =   Since the series in the brackets diverges, so does the series 3 5 1 k k ∞ ∑ =

• Problem. Use any method to determine if the following series

converges or diverges. 1 1k k ∞ ∑ = Solution 2. The series is a p - series, and so by the p - series test it diverges. 3 5 1 k k ∞ ∑ = 3 5 1 k k ∞ ∑ = is a constant multiple of this p - series, so also diverges.

• Problem. Use any method to determine if the following series

converges or diverges. 2 1 2 3 1 k k k ∞ + ∑ + =

• Solution. Again check the divergence test. In this case we have

Thus by the divergence test, the series diverges. 1 1 2 2 1 lim lim 1 0 2 3 3 1 2 k k k k k k + + = = ≠ →∞ →∞ + + 2 1 2 3 1 k k k ∞ + ∑ + =

• Problem. Use any method to determine if the following series

converges or diverges. ln 3 k k k ∞ ∑ =

• Solution. Here the function is integrable and

satisfies the conditions of the integral test. Thus the series behaves exactly as does the improper integral ln ( ) x f x x = ln ln lim 3 3 r x x dx dx x x r ∞ = ∫ ∫ →∞ We pause to evaluate ln . xdx x ∫ Let u = ln x, du = dx/x.

Therefore the series diverges.

[ ]

2 2 ln ln 1 ln( ) ln(3) 2 lim lim ln( ) lim 2 2 2 3 3 3 r r x x r dx dx x x x r r r   ∞  = = = − =∞   ∫ ∫    →∞ →∞ →∞    Then 2 ln 2 1 ln( ) 2 2 x u dx udu x x = = =   ∫ ∫   Continuing with the previous equation , we have

• Problem. Use any method to determine if the following series

converges or diverges. 2 1 k ke k ∞ − ∑ =

• Solution. Here the function is integrable and

satisfies the conditions of the integral test. Thus the series behaves exactly as does the improper integral 2 ( ) x f x xe− = 2 2 lim 1 1 r x x xe dx xe dx r ∞ − − = ∫ ∫ →∞ We pause to evaluate 2 . x xe dx − ∫ Let u = x2, du = 2xdx.

Therefore the series converges (but not to 1/2e). 2 2 2 1 1 1 lim lim 2 2 1 1 r x x r xe dx xe dx e e e r r ∞   − − − − = =− − =   ∫ ∫ →∞ →∞    Then 2 2 1 1 1 2 2 2 x u u x xe dx e du e e − − − − = =− =− ∫ ∫ Continuing with the previous equation , we have

• Problem. Use any method to determine if the following series

converges or diverges. 1 1 1 1 1 1 3 4 5 1 2 k k k ∞ = + + + + + ∑ + + =

• Solution 1. This series is a p - series with p = 1/2, with 2 terms

dropped off. Thus it diverges. Solution 2. We can use the integral test directly. lim lim 2 1 lim 2 1 2 3 1 1 2 2 2 r r dx dx x r x x r r r ∞    = = + = + − =∞ ∫ ∫    + + →∞ →∞ →∞ We see again that the series must diverge.