Proof since the series converges its partial sums s k

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Proof. Since the series converges, its partial sums s k form a convergent sequence { s k } . By Cauchy’s theorem, { s k } is a Cauchy sequence, that is, | s m - s k | → 0 as k, m → ∞ . Taking k = m - 1, we see that a m = s m - s m - 1 0 as m → ∞ . Remark. The converse is FALSE: it may well happen that lim n →∞ a n = 0 but the series n =1 a n diverges. For instance, the series n =1 1 /n diverges. 2
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Theorem. If n =1 a n and n =1 b n are convergent series then n =1 ( a n + b n ) = n =1 a n + n =1 b n and n =1 c a n = c n =1 a n for all c R . Proof . Let s m = m n =1 a n and σ m = m n =1 b n . Then from the algebraic rules for limits it follows that lim m →∞ m X n =1 ( a n + b n ) = lim m →∞ ( s m + σ m ) = lim m →∞ s m + lim m →∞ σ m = X n =1 a n + X n =1 b n , which means that n =1 ( a n + b n ) = n =1 a n + n =1 b n . Similarly, X n =1 c a n = lim m →∞ m X n =1 c a n = lim m →∞ c s n = c lim m →∞ s n = c X n =1 a n . ABSOLUTELY CONVERGENT SERIES Definition. A series n =1 a n is said to be absolutely convergent if the positive series n =1 | a n | converges. Instead of saying that the positive series is absolutely convergent, one often writes n =1 | a n | < . Obviously, a positive series n =1 a n is convergent if and only if it is absolutely convergent (in this case a n = | a n | ). Absolute Convergence Theorem. Every absolutely convergent series is con- vergent. Or: if n =1 | a n | < then n =1 a n converges. Proof. Let s k = k n =1 a n and σ k = k n =1 | a n | . If m > k then, using the triangle inequality for the modules, we obtain | s m - s k | = m X n = k +1 a n 6 m X n = k +1 | a n | = σ m - σ k . Since the series n =1 | a n | converges, the sequence { σ k } converges. Therefore it is a Cauchy sequence, that is, σ m - σ k 0 as m, k → ∞ . The above inequality implies that | s m - s k | → 0 as m, k → ∞ , which means that { s k } is also a Cauchy sequence. Finally, by Cauchy’s theorem, the sequence { s k } converges.
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