# Comparison theorem for series if a n 6 b n for all

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Comparison Theorem for Series. If | a n | 6 b n for all sufficiently large n and the series n =1 b n is convergent then the series n =1 a n is absolutely convergent. Proof. The words “for all sufficiently large n ” mean that | a n | 6 b n for all n > n 0 , where n 0 is some fixed positive integer. Denote σ k = k n =1 | a n | and σ 0 k = k n =1 b n . Then 0 6 ( σ m - σ k ) 6 ( σ 0 m - σ 0 k ) whenever m > k > n 0 . The right hand side of this inequality converges to 0 as m, k → ∞ because the sequence { σ 0 k } is convergent (Cauchy’s theorem). There- fore | σ m - σ k | → 0 as m, k → ∞ and, consequently, { σ k } is a Cauchy sequence. 3

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By Cauchy’s theorem, the sequence { σ k } converges, which means that the series n =1 a n is absolutely convergent. Definition. We say that R 0 f ( x ) d x = c if lim n →∞ (R n 0 f ( x ) d x ) = c . Integral Comparison Theorem. If f ( x ) is a positive and monotonic decreasing function on [1 , ) then the sum n =1 f ( n ) converges if and only if the integral R 1 f ( x ) d x converges. (This is not stating that the sum and integral are equal.) Proof. The theorem follows from Comparison Theorem for Series and “obvious” estimates f ( n ) 6 R n +1 n f ( x ) d x 6 f ( n +1). The “obvious” estimates will be proved in the end of the course. Example. The series n =1 1 /n α converges if α > 1 and diverges if α 6 1. 4
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