Review of series

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Review of series

  1. 1. MATH 1220 Summary of Convergence Tests for Series ∞Let ∑a n=1 n be an infinite series of positive terms. ∞The series ∑a n=1 n converges if and only if the sequence of partial sums, ∞S n = a1 + a 2 + a3 +  a n , converges. NOTE: lim S n = ∑a n n→∞ n=1 ∞Divergence Test: If n → ∞ an ≠ 0 , the series lim ∑a n diverges. n=1 n 1 ∞ n lim = lim =1Example: The series ∑ is divergent since n→ ∞ n2 + 1 n→ ∞ 1+ 1 n =1 n +12 n2This means that the terms of a convergent series must approach zero. That is, if ∑ n aconverges, then lim an = 0. However, lim a n = 0 does not imply convergence. n →∞ n →∞Geometric Series: THIS is our model series A geometric seriesa + ar + ar 2 +  + ar n −1 +  converges for − 1 < r < 1 . an +1 aNote: r = If the series converges, the sum of the series is . an 1 −r n ∞ 7  35 7Example: The series ∑5 8  converges with a = a1 = 8 and r = 8 . The sum of the n=1  series is 35.Integral Test: If f is a continuous, positive, decreasing function on [1, ∞ with ) ∞ f ( n) = an , then the series ∑a n=1 n converges if and only if the improper integral∞∫ f ( x)dx converges.1 ∞Remainder for Integral Test: If ∑a n=1 n converges by the Integral Test, then the ∞remainder after n terms satisfies Rn ≤∫ f ( x)dx n ∞ 1p-series: The series ∑n n=1 p is convergent for p > 1 and diverges otherwise. ∞ ∞ 1 1Examples: The series ∑n n=1 1.001 is convergent but the series ∑n n=1 is divergent.
  2. 2. ∞ ∞Comparison Test: Suppose ∑a n=1 n and ∑b n=1 n are series with positive terms. ∞ ∞(a) If ∑b n=1 n is convergent and an ≤ bn for all n, then n=1 ∑a n converges. ∞ ∞(b) If ∑b n=1 n is divergent and an ≥ bn for all n, then ∑a n=1 n diverges.The Comparison Test requires that you make one of two comparisons: • Compare an unknown series to a LARGER known convergent series (smaller than convergent is convergent) • Compare an unknown series to a SMALLER known divergent series (bigger than divergent is divergent) ∞ ∞ ∞ 3n 3n 1Examples: ∑n n =2 2 > ∑ 2 = 3∑ which is a divergent harmonic series. Since the − 2 n =2 n n =2 noriginal series is larger by comparison, it is divergent. ∞ ∞ 5n 5n 5 ∞ 1We have ∑ 2n 3 + n 2 + 1 n=1 2n 2 n=1 n n =1 < ∑ 3 = ∑ 2 which is a convergent p-series. Since theoriginal series is smaller by comparison, it is convergent. ∞ ∞Limit Comparison Test: Suppose ∑a n=1 n and ∑b n=1 n are series with positive terms. If anlim = c where 0 < c < ∞ , then either both series converge or both series diverge.n →∞ b n(Useful for p-series)Rule of Thumb: To obtain a series for comparison, omit lower order terms in thenumerator and the denominator and then simplify. ∞ ∞ ∞ n 1Examples: For the series ∑ 2 n n =1 n + n + 3 , compare to ∑n 2 =∑ 3 which is a n =1 n =1 n 2convergent p-series. πn + n n ∞ ∞ πn ∞ π For the series ∑ n 2 , compare to ∑ n = ∑  which is a divergent geometric n =1 3 + n n =1 3 n =1  3 series.Alternating Series Test: If the alternating series ∞∑( −1) n −1 bn = b1 − b2 + b3 − b4 + b5 − b6 + n =1satisfies (a) bn > bn +1 and (b) n →∞ bn = 0 , then the series converges. limRemainder: Rn = s − sn ≤ bn +1Absolute convergence simply means that the series converges without alternating (allsigns and terms are positive). ∞ ( −1) nExamples: The series ∑ n =0 n +1 is convergent but not absolutely convergent. (− ) n 1 ∞Alternating p-series: The alternating p-series ∑ p converges for p > 0. n=1 n
  3. 3. ∞ ( −1) n ∞ ( −1) nExamples: The series ∑ n=1 n and the Alternating Harmonic series ∑ n=1 n areconvergent.
  4. 4. ∞ an +1 an +1 limRatio Test: (a) If n →∞ an < 1 then the series ∑a n=1 n lim converges; (b) if n →∞ an >1the series diverges. Otherwise, you must use a different test for convergence.This says that if the series eventually behaves like a convergent (divergent) geometricseries, it converges (diverges). If this limit is one, the test is inconclusive and a differenttest is required. Specifically, the Ratio Test does not work for p-series.POWER SERIES ∞Radius of Convergence: The radius of convergence for a power series ∑c n =0 n ( x − a) n cnis R = n →∞ lim . The center of the series is x = a. The series converges on the open cn +1interval ( a − R, a + R ) and may converge at the endpoints. You must test each seriesthat results at the endpoints of the interval separately for convergence. ∞ ( x + 2) nExample: The series ∑ (n +1) n =0 2 is convergent on [-3,-1] but the series ∞ (−1) n ( x − 3) n∑n =0 5n n +1 is convergent on (-2,8].Taylor Series: If f has a power series expansion centered at x = a, then the power series ∞ f ( n ) (a)is given by f ( x) = ∑ ( x − a) n . Use the Ratio Test to determine the interval of n =0 n!convergence. (n + )Taylor’s Inequality (Remainder): If f 1 ( x ) ≤ M for x − ≤d , then a M n +1 Rn ( x) ≤ x −a for x −a ≤d . Note that d < R (the radius of convergence) ( n +1)! ( n+ ) 1and think of M as the maximum value of f ( x ) on the interval [x,a] or [a,x].

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