# Download Elementary Number Theory, 5th Edition by David M. Burton PDF

By David M. Burton

This article offers an easy account of classical quantity idea, in addition to a few of the old historical past within which the topic developed. it's meant to be used in a one-semester, undergraduate quantity thought direction taken essentially through arithmetic majors and scholars getting ready to be secondary institution academics. even if the textual content used to be written with this viewers in brain, only a few formal necessities are required. a lot of the textual content may be learn via scholars with a valid history in highschool arithmetic.

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Extra info for Elementary Number Theory, 5th Edition

Example text

71. A Fejer limit. Let f ∈ L1 [0, 1] and let g : [0, 1] → R be a continuous function. Prove that 1 1 f (x)g({nx})dx = lim n→∞ 0 0 f (x)dx 1 g(x)dx, 0 where {a} = a − a denotes the fractional part of a. 72. Let f , g : [0, 1] × [0, 1] → R be two continuous functions. Prove that 1 1 lim n→∞ 0 0 f ({nx} , {ny})g(x, y)dxdy = 1 0 1 0 f (x, y)dxdy · 1 1 g(x, y)dxdy, 0 0 where {a} = a − a denotes the fractional part of a. 73. Let f : [−1, 1] × [−1, 1] → R be a continuous function. Prove that 1 lim t→∞ −1 f (sin(xt), cos(xt)) dx = 1 π 2π 0 f (sin x, cos x)dx and 1 1 lim t→∞ −1 −1 f (sin(xt), cos(yt)) dxdy = 1 π2 2π 0 2π 0 f (sin x, cos y)dxdy.

13. It is worth mentioning the history of this problem. It appears that the case s = 1 is an old problem which Bromwich attributed to Joseph Wolstenholme (1829–1891) (see [20, Problem 19, p. 528]). The problem, as Bromwich records it, states that lim n→∞ 1 n n + 2 n n + ···+ n−1 n n = 1 , e−1 which we call the Wolstenholme limit. In 1982, I. J. Schoenberg proposed the problem of proving that limn→∞ Sn = 1/(e − 1) where Sn = (n + 1)−n ∑nk=1 kn (see [114]). However, it appears that this problem appeared previously in Knopp’s book [76, p.

1, 1 + 1n , . . , k − 2, k − 2 + 1 n−1 n , . . , k − 2 + n , k − 1. Since f is Riemann integrable we obtain that √ √ √ p n + p n + 1 + · · · + p kn √ = lim n→∞ npn k−1 0 √ p 1 + xdx = p · k(p+1)/p − 1 . 15), we get that xn + xn+1 + · · · + xkn p L−ε · k(p+1)/p − 1 ≤ lim · n→∞ L+ε p+1 nxn ≤ L+ε p · k(p+1)/p − 1 . · L−ε p+1 Since ε > 0 is arbitrary taken the result follows. 19. The problem, which is recorded as a lemma (see [91, Lemma, p. 434]), is used for introducing a new approach for determining asymptotic evaluations of products of the form f (n) = c1 c2 · · · cn .