be [prove] a great boon to [for]

本文探讨了资产证券化——即通过资本市场将资产从商业银行资产负债表转移到投资者手中的过程——如何成为投资银行的一大福音。

boon

[bu:n]
n.

[古]请求

赐物; 恩惠; 福利, 方便

麻杆; 麻骨

羊毛皮夹杂的植物性杂质

 

be [prove] a great boon to [for]
(成为)对...极可感谢的[有用的]东西收起更多词典


Securitization – the shift of assets from commercial bank balance sheets to investors via the capital markets – has been a particular boon to investment banks.

The Cauchy principle of convergence states that a sequence converges if and only if it is a Cauchy sequence. To prove this principle, we need to show two things: 1. If a sequence converges, then it is a Cauchy sequence 2. If a sequence is a Cauchy sequence, then it converges Proof of 1: Let {an} be a convergent sequence. Then there exists a limit L such that for any ε > 0, there exists an N such that for all n > N, |an - L| < ε. Now let ε > 0 be arbitrary. We want to show that there exists an N such that for all n, m > N, |an - am| < ε. Since {an} converges to L, we can choose N1 such that for all n > N1, |an - L| < ε/2. Similarly, we can choose N2 such that for all m > N2, |am - L| < ε/2. Let N = max{N1, N2}. Then for all n, m > N, |an - am| = |an - L + L - am| ≤ |an - L| + |L - am| < ε/2 + ε/2 = ε Thus, {an} is a Cauchy sequence. Proof of 2: Let {an} be a Cauchy sequence. Then for any ε > 0, there exists an N such that for all n, m > N, |an - am| < ε. Since {an} is Cauchy, it is also bounded. Let M be an upper bound on {an}. Now consider the set {an : n ≥ N}. This set is bounded above by M and bounded below by {an - ε : n ≥ N}, since for any n ≥ N, an - ε ≤ an ≤ an + ε Therefore, by the completeness axiom of the real numbers, this set has a supremum, which we will call L. We claim that {an} converges to L. To prove this, let ε > 0 be arbitrary. Since {an} is Cauchy, there exists an N such that for all n, m > N, |an - am| < ε/2. Choose N such that N > N1 and N > N2, where N1 and N2 are the indices corresponding to ε/2 and ε/4, respectively. Then for all n > N, |an - L| ≤ |an - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4 Similarly, for all m > N, |am - L| ≤ |am - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4 Therefore, for all n > N, |an - L| ≤ |an - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4 and for all m > N, |am - L| ≤ |am - aN| + |aN - L| < ε/2 + ε/4 = 3ε/4 Thus, for all n > N, |an - L| < 3ε/4 and for all m > N, |am - L| < 3ε/4 Therefore, for all n > N, |an - L| + |L - am| < ε/2 + ε/2 = ε which shows that {an} converges to L.
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