Series Summary P General strategy for testing convergence of a series an. 1. Does limn!1 an = 0? If not, then the series diverges and you're done. 2. Does the series have negative terms? (For example, is it an alternating series?) Then test P whether janj converges. If it does not, use the alternating series test. (Note: If you use the ratio or root test and are not in the L = 1 case, you are done and do not need to use the AST.) Alternating Series Test P n If the series (−1) bn, with bn ≥ 0 satisfies (a) limn!1 bn = 0, and (b) bn+1 ≤ bn for n sufficiently large, then it converges. 3. To test for convergence of a series with positive terms, we have the following tests. Direct Comparison Test Suppose an ≤ bn for all n. P P (a) If bn converges, then an converges P P (b) If an diverges, then bn diverges. Limit Comparison Test P P an Consider the series an and bn. Suppose lim = c: n!1 bn If c is finite and c > 0, then either both series converge or both series diverge. 4. The following tests can be used for any series with both positive and negative terms, but are not always conclusive. an+1 Ratio Test Let lim = L. n!1 an P (a) If L < 1, then an is absolutely convergent (and hence convergent) P (b) If L > 1 (or infinite), then an diverges. (c) If L = 1, the Ratio Test is inconclusive, and you should use a different test. 1 pn Root Test Let lim janj = L. n!1 P (a) If L < 1, then an is absolutely convergent (and hence convergent) P (b) If L > 1 (or infinite), then an diverges. (c) If L = 1, the Root Test is inconclusive, and you should use a different test. Here are some of the standard types of series you might encounter. 1. Type: Rational functions (including roots of polynomials) 1 p 1 X n4 + 3n + 1 X 1 Ex. converges because it behaves like , which is a convergent p- n6 n4 n=1 n=1 series. Test: Limit Comparison Test 2. Type: Fraction of sums involving n-th powers 1 1 1 X 3n X n2 + 3n X 3n Ex. or both converge because they both behave like , n − 4n n + 4n 4n n=1 n=1 n=1 which is a convergent geometric series. Test: Limit Comparison Test 3. Type: Combinations (meaning multiplication and division) of factorials, n-th powers, and constant powers of n. 1 1 1 1 X n3 X n! X n10 X 1 · 3 · 5 ··· (2n + 1) Ex. or or or 4n 3n+1 (3n)! n! n=1 n=1 n=1 n=1 Test: Ratio Test 4. Type: Compositions involving either trigonometric functions, exponentials, or logarithms 1 1 X 2n + 1 X 2n + 1 Ex. ln or sin 3n + 1 3n + 1 n=1 n=1 Test: Try the test for divergence 5. Fractions of functions raised to the n-th power 1 n 1 X n + 1 X sinn(1=n) Ex. or 2n + 1 n2n n=1 n=1 Test: Root Test 2 6. Type: Trigonometric functions, logarithms, or exponentials over a power of n or an n-th power 1 1 1 X ln n X arctan n X 31=n Ex. or or n 3n 2n n=1 n=1 n=1 Test: Try the comparison test (usually not the limit comparison test) Remarks There are often many ways to test for convergence/divergence. Notice, for example, that I did not mention the integral test. There are many cases when the integral test is applicable, and if you're stuck and can integrate the function, then this is a good backup. Because it is somewhat messy, I tend to avoid using it. You may be asked, however, to prove that a p-series converges using the integral test. So you should know how to apply it. Here are some examples that are a little bit ambiguous, but should not be difficult. 1 X n2 1. n + 2n n=1 Here you can use the ratio test off the bat, or you can notice that the n in the denominator 1 X n2 is meaningless in the limit, and so it should behave like , which converges by the ratio 2n n=1 1 X 1 test. Do not try to compare this to . The n2 in the numerator does not behave like 1 2n n=1 in the limit, but the n + 2n in the denominator does behave like 2n in the limit. 1 X ln n 2. n2 n=1 In this example, it does not help to use the fact that ln n > 1 or that ln n < n, because the p comparison test gives no information in both cases. You can, however, say that ln n < n p eventually. This can be rigorously proved by showing that ln n − n is eventually decreasing using its derivative. Alternatively, you can use the integral test. 1 X en 3. n n=1 This one is extremely easy, because the test for divergence works here. However, one can also use the fact that en > 1 and use the divergence of the harmonic series, which is an even easier test. It's even possible to do it using the ratio test, but that's overkill. 3.