Testing for Convergence or Divergence of a Series PDF

Testing for Convergence or Divergence of a Series Many of the series you come across will fall into one of several basic types. Recognizing these types will help you decide which tests or strategies will be most useful in finding whether a series is convergent or divergent. p-Series...

Testing for Convergence or Divergence of a Series Many of the series you come across will fall into one of several basic types. Recognizing these types will help you decide which tests or strategies will be most useful in finding whether a series is convergent or divergent. p-Series Geometric Series ∞ ∞ 1 ∑ n =1 n p is… convergent if p > 1 ∑ ar n −1 is… convergent if r < 1 n =1 divergent if p ≤ 1 divergent if r ≥ 1 If a n has a form that is similar to one of the above, see whether you can use the comparison test: ∞ 1 Example: ∑ 2 Comparison Test n =1 n + n (Warning! This only works if a n and bn are always positive.) 1 Pick bn = 2 (p-series) (i) If a n ≤ bn for all n, and ∑ bn is convergent, then ∑ a n is convergent. n (ii) If a n ≥ bn for all n, and ∑ bn is divergent, then ∑ a n is divergent. 1 1 an = 2 ≤ 2 , and n +n n ∞ 1 ∑ converges, so by Consider a series ∑ bn so that the ratio 2 n =1 n ∞ ∞ 1 Example: ∑ n a n bn cancels the dominant terms in (i), ∑ 2 1 converges. n =1 2 − 1 n =1 n + n 1 the numerator and denominator of a n , Pick bn = n (geometric) as in the example to the left. If you 2 a 1 2 n know whether ∑ bn converges or not, lim n = lim n n →∞ b n →∞ 2 − 1 1 try using the limit comparison test. n 1 = lim =1> 0 n →∞ 1 − 1 2 n Limit Comparison Test ∞ 1 (Warning! This only works if a n and bn are always positive.) ∑ converges, so a If lim n = c > 0 (and c is finite), then ∑ a n and ∑ bn either both n n =1 2 ∞ n →∞ b 1 ∑ n converges. n =1 2 − 1 n converge or both diverge. Some series will “obviously” not n2 −1 ∞ converge—recognizing these can save you a lot of time and guesswork. Example: ∑ n =1 n + n 2 1 Test for Divergence 1− n −1 2 n2 = 1 ≠ 0 ∞ lim a n = lim 2 = lim If lim a n ≠ 0 , then n →∞ ∑a n =1 n is divergent. n →∞ n →∞ n + n n →∞ 1+ 1 n ∞ n −1 2 so ∑ 2 is divergent. n =1 n + n Testing for Convergence or Divergence of a Series (continued) If a n can be written as a function with a “nice” ∞ 1 integral, the integral test may prove useful: Example: ∑n n =1 2 +1 Integral Test 1 f ( x) = is continuous, If f (n) = a n for all n and f (x) is continuous, x +1 2 positive, and decreasing on [1, ∞ ) , then: positive, and decreasing on [1, ∞ ). ∞ ∞ t ∞ 1 1 If ∫ f ( x)dx converges, then ∑a n converges. ∫1 x 2 + 1 dx = lim t →∞ ∫ 1 x +1 2 dx 1 n =1 π ] ∞ t ∞ = lim tan −1 x = lim tan −1 t − If ∫ f ( x)dx is divergent, then ∑ an is divergent. n =1 t →∞ 1 t →∞ 4 1 π π π ∞ 1 = 2 4 4 − = , so ∑n n =1 2 +1 is ∞ 2 n ∑ (−1) n −1 convergent. Example: n =1n +1 3 1 1 1 1 (i) = n + 2 , so = n +1+ > n +1 Alternating Series Test bn n bn +1 (n + 1) 2 If (i) bn +1 ≤ bn for all n and (ii) 1 1 1 1 ≥ n+ 2 = , so ≥ , so bn +1 ≤ bn ∞ n bn bn +1 bn lim bn = 0 , then n →∞ ∑ (−1) n =1 n −1 bn is n2 1 (ii) lim = lim =0 convergent. n →∞ n + 1 3 n →∞ n + 1 n 2 ∞ n2 So ∑ (−1) n−1 n =1 n3 + 1 is convergent. The following 2 tests prove convergence, but also prove the stronger fact that ∑a n converges (absolute convergence). Ratio Test ∞ a Example: ∑e −n n! If lim n +1 < 1 , then ∑a n is absolutely convergent. n =1 n →∞ a n a n +1 e − n −1 (n + 1)! lim = lim a n +1 a n →∞ a n →∞ e − n n! If lim n →∞ a > 1 or lim n +1 = ∞ , then n →∞ a ∑a n is divergent. n = e −1 lim n + 1 = ∞ , so n n n →∞ a ∞ If lim n +1 = 1 , use another test. n →∞ a n ∑e n =1 −n n! is divergent. ∞ nn When a n contains factorials and/or powers of constants, Example: ∑ n =1 3 1+ 3 n as in the above example, the ratio test is often useful. nn n lim 1+3n = lim 1 n 3 n Root Test ∑a n →∞ 3 n → ∞ 3 3 If lim n a n < 1 , then n is absolutely convergent. n →∞ 1 n = lim 1 n = ∞ a n +1 27 n→∞ 3 ∞ If lim n a n > 1 or lim n →∞ n →∞ a = ∞ , then ∑a n is divergent. nn n So ∑ 1+3n is divergent. n =1 3 If lim n a n = 1 , use another test. n →∞