# Series Test Proofs - Math 142 Series Test Proofs...

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Math 142: Series Test Proofs Theorem: (The Monotone Convergence Theorem) If a n is a decreasing sequence that is bounded below, then it converges. Similarly, is a n is increasing and bounded above, then it converges. Proof: Suppose a n is decreasing and bounded below. Let > 0, and consider the greatest lower bound L of the sequence (this exists by the completeness axiom). Then by definition of greatest lower bound, L + is not a lower bound of a n . Let N be the smallest value such that a N < L + . Then since a n is decreasing, we know that a n < L + for all n N . Finally, this says that a n - L < for all n N , and since L is a lower bound of a n , we know that a n - L 0. Thus, | a n - L | < , so lim n →∞ a n = L by definition. Suppose a n is increasing and bounded below. The proof is identical, except this time we let L be the least upper bound of the sequence, note that L - is not an upper bound of a n , and find an N such that a n > L - for all n N . Since L - a n 0, we get that | a n - L | < . Theorem: (Geometric Series) The geometric series X n =1 ar n - 1 converges to a 1 - r when | r | < 1 and diverges when | r | ≥ 1. Proof: First, we’ll get an expression for s n : s n = a + ar + ar 2 + ar 3 + ... + ar n - 1 rs n = ar + ar 2 + ar 3 + ... + ar n - 1 + ar n Subtracting these two equations, we get that s n - rs n = a - ar n , so s n (1 - r ) = a (1 - r n ), and finally, we get an expression for s n : a (1 - r n ) 1 - r . We now proceed to take the limit of s n . If | r | < 1, lim n →∞ a (1 - r n ) 1 - r = a (1 - 0) 1 - r = a 1 - r , so it converges to a 1 - r . If | r | > 1, lim n →∞ r n diverges, so s n diverges and hence the series diverges. If r = 1, then the series is simply X n =1 a = a + a + a + ... , which diverges. If r = - 1, then the series is simply X n =1 a ( - 1) n - 1 = a - a + a - a + ... , which diverges. Theorem: (The Divergence Test) If lim n →∞ a n 6 = 0 or does not exist, then X n =1 a n diverges. Proof: We’ll prove the contrapositive: If the series X n =1 a n is convergent, then lim n →∞ a n = 0. Notice that a n = s n - s n - 1 , where s n is the n th partial sum of X n =1 a n . Since X n =1 a n converges, s n s . Clearly, this means that s n - 1 s as well. So, lim n →∞ a n = lim n →∞ ( s n - s n - 1 ) = lim n →∞ s n - lim n →∞ s n - 1 = s - s = 0. 1
Theorem: (Constant Multiples of Series) If X n =1 a n converges, then X n =1 ca n converges to c X n =1 a n . If X n =1 a n diverges, then X n =1 ca n diverges. Proof: Let s n be the partial sums of X n =1 a n . If X n =1 a n converges, then say s n s . The n th partial sum of X n =1 ca n is ca 1 + ca 2 + ... + ca n = c ( a 1 + a 2 + ... + a n ) = cs n . So, lim n →∞ cs n = c lim n →∞ s n = cs . If X n =1 a n diverges, then lim n →∞ s n diverges. Thus, lim n →∞ cs n diverges, so X n =1 ca n diverges.