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 • < ∑ p is… convergent if p 1 ∑ ar is… convergent if r 1 n=1 n n=1 • divergent if p ≤ 1 • divergent if r ≥ 1

If an 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 and b are always positive.) 1 n n = Pick bn 2 (p-series) (i) If an ≤ bn for all n, and ∑bn is convergent, then ∑ an is convergent. n 1 1 ≥ = ≤ (ii) If an bn for all n, and ∑bn is divergent, then ∑ an is divergent. an 2 2 , and n + n n ∞ 1 ∑ 2 converges, so by n=1 n ∞ Consider a series ∑bn so that the ratio 1 ∞ Example: 1 ∑ n an bn cancels the dominant terms in (i), converges. = 2 −1 ∑ 2 n 1 n=1 n + n 1 the numerator and denominator of an , Pickb = (geometric) n 2n as in the example to the left. If you n a 1 2 know whether ∑bn converges or not, lim n = lim n→∞ n→∞ n try using the limit comparison test. bn 2 −1 1 1 = lim = 1 > 0 n→∞ 1−1 2n Limit Comparison Test ∞ 1 (Warning! This only works if an and bn are always positive.) ∑ n converges, so n=1 2 an If lim = c > 0 (and c is finite), then an and bn either both ∞ n→∞ ∑ ∑ 1 bn ∑ n converges. n=1 2 −1 converge or both diverge.

Some series will “obviously” not 2 converge—recognizing these can save ∞ n −1 Example: ∑ 2 you a lot of time and guesswork. n=1 n + n 1 − Test for Divergence 2 1 2 ∞ n −1 n lim an = lim = lim = 1 ≠ 0 n→∞ n→∞ 2 n→∞ If lim an ≠ 0 , then an is divergent. n + n 1 n→∞ ∑ 1+ n=1 n ∞ n 2 −1 so ∑ 2 is divergent. n=1 n + n Testing for Convergence or Divergence of a Series (continued)

If a can be written as a function with a “nice” n ∞ 1 integral, the integral test may prove useful: Example: ∑ 2 n=1 n +1 1 Integral Test = f (x) 2 is continuous, If f (n) = an for all n and f (x) is continuous, x +1 positive, and decreasing on [1,∞), then: positive, and decreasing on [1,∞). ∞ t ∞ ∞ 1 1 = If f (x)dx converges, then a converges. 2 dx lim 2 dx ∫ ∑ n ∫ x +1 t→∞ ∫ x +1 1 n=1 1 1 ∞ t π ∞ = limtan −1 x] = lim tan −1 t − If ∫ f (x)dx is divergent, then ∑ an is divergent. t→∞ 1 t→∞ 4 = 1 n 1 π π π ∞ 1 = − = , so ∑ 2 is 2 4 4 n=1 n +1 ∞ n 2 − n−1 convergent. Example: ∑( 1) 3 n=1 n +1 1 1 1 1 = + = + + > + (i) n 2 , so n 1 2 n 1 Alternating Series Test bn n bn+1 (n +1) If (i) b + ≤ b for all n and (ii) 1 1 1 1 n 1 n ≥ + = ≥ ≤ ∞ n 2 , so , so bn+1 bn n−1 n b b + b limbn = 0 , then (−1) bn is n n 1 n n→∞ ∑ n=1 n 2 1 (ii) lim = lim = 0 convergent. n→∞ n3 +1 n→∞ n +1 n 2 ∞ n 2 − n−1 So ∑( 1) 3 is convergent. n=1 n +1

The following 2 tests prove convergence, but also prove the stronger fact that ∑ an converges (absolute convergence).

Ratio Test ∞ Example: ∑e −n n! an+1 n=1 If lim < 1, then an is absolutely convergent. n→∞ ∑ −n−1 an a e (n +1)! lim n+1 = lim n→∞ n→∞ −n an+1 an+1 an e n! If lim > 1 or lim = ∞ , then an is divergent. n→∞ n→∞ ∑ −1 an an = e lim n +1 = ∞ , so n→∞ a ∞ If lim n+1 = 1, use another test. −n n→∞ ∑e n! is divergent. an n=1

∞ n n When an contains factorials and/or powers of constants, Example: ∑ 1+3n n=1 3 as in the above example, the ratio test is often useful. n n n lim n = lim Root Test n→∞ 1+3n n→∞ 1 n 3 3 3 3 n If lim an < 1, then an is absolutely convergent. n→∞ ∑ 1 n = lim = ∞ a 27 n→∞ 31 n If lim n a > 1 or lim n+1 = ∞ , then a is divergent. →∞ n →∞ ∑ n ∞ n n n a n n So ∑ 1+3n is divergent. n = If lim an = 1, use another test. n 1 3 n→∞