Tend to Infinity

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Tend to Infinity

Describing a variable that, for any reason, becomes extremely large.
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and also for a sequence of values of r tending to infinity we get that
Assuming that this conjecture holds, we can prove a lower bound of 0.3718 and an upper bound of 1.8310 for the expected value of a minimum cost hyperassignment in [G.sub.2,2n] for the exponential edge cost distribution and for vertex numbers tending to infinity. To achieve this, we first use a combinatorial argument to represent the bounds in terms of bounds for random assignments.
Since [T.sup.-1.sub.h] (r) is an increasing function of r, it follows from Lemma 1 that, for a sequence of values of r tending to infinity,
where j and k are tending to infinity independent of each other.
Let [LAMBDA] = ([[lambda].sub.n]) be a non-decreasing sequence of positive real numbers tending to infinity and [[lambda].sub.1] = 1 and [[lambda].sub.n+1] [less than or equal to] [[lambda].sub.n] + 1, for all n [member of] N.
And the consecutive repair times maybe increasing and tending to infinity. Let [Z.sub.n] be the repair time after nth failure.
The double sequence [[lambda].sub.2] = {([[beta].sub.r], [[mu].sub.s])} is called double [[lambda].sub.2] sequence if there exist two non-decreasinig sequences of positive numbers tending to infinity such that [[beta].sub.r+1] [less than or equal to] [[beta].sub.r] + l, [[beta].sub.1] = l and [[mu].sub.s+1] [less than or equal to] [[mu].sub.s] + l, [[mu].sub.1] = l.
Also let [([[mu].sub.n]).sup.[infinity].sub.n=0] be a nondecreasing sequence of positive reals tending to infinity. Furthermore let [LAMBDA] = [([[lambda].sub.n]).sup.[infinity].sub.n=0] be an exponentially bounded sequence, and [([[lambda].sub.n(v)]).sup.[infinity].sub.v=0] an associated subsequence with [[lambda].sub.n(0)] = [[lambda].sub.0].
In particular, we still cannot show that deviations of size [n.sup.l/2][omega](n) have probability tending to 0 for [omega](n) tending to infinity arbitrarily slowly.
If the spectrum of the preconditioned matrix is bounded ([[lambda].sub.max] [less than or equal to] [alpha] < [infinity], [alpha] independent of n), then [[lambda].sub.min] [right arrow] [infinity] and [beta](n) eigenvalues tend to zero, with [beta](n) tending to infinity when n tends to infinity.
Any negative value for [C.sub.t] will lead to the sequence of [C.sub.t]'s converging to zero or tending to infinity. This will violate the transversality condition above and thus provide a contradiction.
Now from (29) and in view of (31), for a sequence of values of r tending to infinity we get that