Notes 10.4.Pptx

10.4

Radius of Convergence

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall What you’ll learn about… … and why n Convergence It is important to develop a n nth-Term Test strategy for finding the interval of convergence of a n Comparing Nonnegative Series power series and to obtain n Ratio Test some tests that can be used n Endpoint Convergence to determine convergence of a series.

EQ: hhh

∞ n There are three possibilities for ∑ cn (x − a) with respect to convergence: n=0 1. There is a positive number R such that the series diverges for x − a > R but converges for x − a < R. The series may or may not converge at either of the endpoints x = a − R amd x = a + R. 2. The series converges for every x (R = ∞). 3. The series converges at x = a and diverges elsewhere (R = 0).

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∞

andiverges if liman fails to exist or is different from zero. ∑ n→∞ n =1

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∞ 2 Example: Show that ∑ k diverges by the nth- k =0 term test for divergence.

2 € limk = ∞, so the series diverges. k →∞

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∞ 1 1 n! n! lim = ∑ lim n→∞ ⎛ 1 ⎞ 2 Diverges 2n!+1 n→∞ 2 +⎜ ⎟ n=1 2n!+1 ⎝ n!⎠

∞ 1 1 ∑ lim = 0 Converges n→∞ n=1 n n

a na+1 LetLet ∑ aa bbee a a s serieseries w withith p positiveositive t eterms,rms, a nandd w withith l ilimm n+=1 L.L. ∑ n n n→∞ = n→∞a nan Then, (a) the series converges if L < 1, € (b) the series diverges if L > 1, (c) the test is inconclusive if L = 1.

∞ n ∞ 22 Determine the convergence or divergence of the series . ∑ nn n=0 n =033 ++1 Use the Ratio Test: 2n+1 n 1 n a n+1 ⎛ 2 + ⎞⎛ 3 +1⎞ lim n+1 = lim 3 +1 = lim € n→∞ n→∞ n n→∞⎜ n+1 ⎟⎜ n ⎟ an 2 ⎝ 3 +1⎠⎝ 2 ⎠ 3n +1 ⎛ 3n +1 ⎞ = lim2 n ⎜ n+1 ⎟ →∞ ⎝ 3 +1⎠ € ⎛ 1 ⎞ 1 ⎜ + n ⎟ = lim2⎜ 3 ⎟ n→∞ 1 € ⎜ 3+ ⎟ ⎝ 3n ⎠ 2 2 = The series converges because <1. 3 3

€ € Example Determining Convergence of a Series

∞ n ∞ 22k! Determine the convergence or divergence of the series . . ∑ nn n=0 2k +1 ! nk =013(3 ++1 ) Use the Ratio Test: ⎛ k +1 ! 2k +1 !⎞ an+1 ( ) ( ) lim = lim⎜ • ⎟ €€ n→∞ a k →∞⎜ k! ⎟ n ⎝ (2k + 3)! ⎠ ⎛ k +1 ⎞ = lim⎜ ( ) ⎟ k ⎜ ⎟ →∞⎝ (2k + 3)(2k + 2)⎠ ⎛ k +1 ⎞ = lim k ⎜ 2 ⎟ →∞⎝ 4k +10k + 6⎠

€ = 0. The series converges because 0 <1.

€ € Example Determining Convergence of a Series ∞ k ∞ 3n ∞ 22 . Determine the convergence or divergence of the series ∑ 3 .k ∑ nn n=0 k 2 nk ==0133 ++1 Use the Ratio Test: € ⎛ k +1 3 k ⎞ ⎛ 3€ ⎞ ⎜ 3 k 2 ⎟ ⎜ 3k ⎟ 3 lim 3 • k = lim 3 = k →∞⎜ k +1 2k +1 3 ⎟ k →∞⎜ k +1 2⎟ 2 ⎝ ( ) ⎠ ⎝ ( ) ⎠

€

3 The series diverges because >1. 2

€ Example Determining Convergence of a Series ∞∞ k ∞ 32n k ∞ k22 e− .. Determine the convergence or divergence of the series ∑∑ 3 .k ∑ nn n=0 k 2 nkk ===01313 ++1 Use the Ratio Test: € €€ 2 2 ⎛ k 1 e−(k +1) ⎞ ⎛ k 1 ⎞ ⎜ ( + ) ⎟ ⎜ ( + ) ⎟ 1 lim 2 = lim 2 = k →∞⎜ k e−k ⎟ k →∞⎜ k e ⎟ e ⎝ ( ) ⎠ ⎝ ⎠

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1 The series converges because <1. e

€ Example Determining Convergence of a Series ∞∞ kk ∞ 32n2 k ∞ k22 e− ... Determine the convergence or divergence of the series ∑∑ k3 .k ∑ nn n=0 k3 2+1 nkk ===01313 ++1 Use the Ratio Test: k k ⎛ k +1 k ⎞ ⎛ 2 3 +1 ⎞ € 2 3 + 2 2 3 +1 ⎜ ( )⎟ €€ ( ) lim⎜ k 1 • k ⎟ = lim k 1 = lim k →∞⎝ 3 + +1 2 ⎠ k →∞⎜ 3 + +1 ⎟ k →∞ 3 3k +1 ⎝ ⎠ ( )

k 2(ln3)(3 ) 2 Apply L'Hopital's rule: lim = € k →∞ 3 ln3 3k 3 ( )( )

2 The€ series converges because <1. 3

€ Assignment: p. 515 #29 – 43

Let ∑an be a series with no negative terms.

(a) ∑an converges if there is a convergent series ∑cn with

an ≥ cn for all n > N, for some integer N.

(b) ∑an diverges if there is a divergent series ∑d n with

an ≥ d n for all n > N, for some integer N.

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€ Remind you of anything?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall Slide 9- 14 Example Proving Convergence by Comparison ∞ 2n x2 n ∞ x ∑ 2 Prove that ∑ 2 converges for all real x. nn=00 n! ((n!)) ∞ 2n x2 n ∞ x Let x be any real number. The series ∑∑ 22 has no negative terms. nn=00 nn!! (( )) n n 2 € 2 n ∞ n 22nn 22n x 2 x2 xx xx x ∞ (x ) For any n, ( ) . Recognize ( ) as the Taylor series n, 22 ≤≤ = . ∑∑ n! n! n! n=0 nn!! (n!) n! n! n =0 2 ( ) 2 xx2 xx2 for e ,, which we know co€nv erges to e for all real numbers. Since ∞ 22n n 2 2 ∞ x x x the e series dominates ∑∑ 22 term by term, the latter series must also n=0 n=0(nn!!) € € ( ) € converge for all real numb€er s by the Direct Comparison Test. € €

∞ n ∞ nnxx Find the radius of convergence of .. ∑ n n=0 n =01100

Check for absolute convergence using the Ratio Test. n+1 n+1 n n an+1 (nn +11) xx € an+1 ( + ) 1010 limlim ==lilimm ⋅⋅ n→n∞→∞ nn→→∞∞ nn++11 n n aan 1010 nnx x n ⎛ n +1⎞⎛ x ⎞ x = lim ⎛ n +1⎞ x = x =n→lim∞ ⎜⎜ ⎟⎟⎜ ⎟ = n→∞⎝ n ⎠⎜10 10⎟ 1100 ⎝ ⎠⎝ ⎠ € x Setting < 1, we see that the series converges absolutely for −10 < x < 10. 10 T€Theh e sserieseries willdiv econvergerges for whenx > 10 |x a|/10nd

€ Absolute Convergence

If the series ∑ an of absolute values converges, then ∑ an converges absolutely.

If a converges, then a converges. ∑ n ∑ n

∞ nn ∞ ((sinsin xx)) Show that ∑ converges for all x. ∑n=0 n =0 nn!!

nn ∞ ∞ ssinin xx Let x be any real number. The series ∑ has no negative terms, ∑n=0 n =0 nn! € ∞ ∞ 1 and it is term-by-term less than or equal to the series ∑∑ , which converges nn ==0 nn!! n n ∞ ∞ ssinin x € to e. Therefore, ∑ converges by direct comparison. n=0 n =0 nn! ∞ nn € ∞ ((sinsin xx)) Since ∑ converges absolutely, it converges. ∑n 0 = n =0 nn!! €

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