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By Joseph R. Lee

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3. Find several partial sums of a rearrangement of 1 — \ + % — I + · · · that are alternately greater than and less than 10. ) //. 4 assuming only one of the series is absolutely convergent and the other just convergent. Criticize the following "proof": let + 5 " ~ " " 6 + T — 8" + "9 ~ S = 1 —2 + è ~ 4 IQ 2 ° _ ~ t"^ = 1 2 1 ~ I "3 1 4 1 1 "I ~6 2~T~ 5" 1 8" I T I I '·· . . 4"Γ9"Η~'·· = S. Let \/2 V3 Λ/Ϊ y/5 and form the Cauchy product of S by itself. Show that the resulting series diverges. 3 hold for infinite sums.

1 completeness of the real number system A rational number can be written as the quotient of two integers p/q, where q ^ 0. The decimal expansion of a real number is actually a geometric series, finite or infinite, whose terms are of the form ak10k, where k = 0, ± 1 , ± 2 , . . 3333... = 3-10" 1 + 3 - 1 0 " 2 + · · · . It is obvious, then, that any terminating decimal is rational; and a repeating decimal may be treated as an infinite series or may be shown to be rational as in the following example.

Then ΣΓ=ι ( — 1 ) *+1α* converges. 2n+l = «1 — («2 — dz) — ( α 4 — α 5 ) — · · · — ( α 2 η — «2n+l) . Then Ain is monotone nondecreasing and A2n+i is monotone nonincreasing. 4; and since lim^«, a2w+i = 0, these limits must be the same, say, A. Then, lim^«, An = A. 8 COROLLARY With the hypotheses stated above, A = An + Rn with limn-oo An = A. Then | Än | < αη+ι· Proof let If n is odd, let n = 2Jfc - 1. Then 0 < A2k-i — A < A2k-i — A2k = a2k. If n is even, let n = 2k. Then 0 < A — A2k < A2k+1 — A2k = a2k+i.

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