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For "The Haruhi Problem" specifically (the case for 14 symbols), the current lower and upper bound are 93,884,313,611 and 93,924,230,411, respectively. [3] This means that watching the series in every possible order would require about 4.3 million years.
The set S = {42} has 42 as both an upper bound and a lower bound; all other numbers are either an upper bound or a lower bound for that S. Every subset of the natural numbers has a lower bound since the natural numbers have a least element (0 or 1, depending on convention). An infinite subset of the natural numbers cannot be bounded from above.
The main objective of interval arithmetic is to provide a simple way of calculating upper and lower bounds of a function's range in one or more variables. These endpoints are not necessarily the true supremum or infimum of a range since the precise calculation of those values can be difficult or impossible; the bounds only need to contain the function's range as a subset.
In graph theory, a bound graph expresses which pairs of elements of some partially ordered set have an upper bound.Rigorously, any graph G is a bound graph if there exists a partial order ≤ on the vertices of G with the property that for any vertices u and v of G, uv is an edge of G if and only if u ≠ v and there is a vertex w such that u ≤ w and v ≤ w.
Stirling's formula is in fact the first approximation to the following series (now called the Stirling series): [6]! (+ + +). An explicit formula for the coefficients in this series was given by G. Nemes. [ 7 ]
Therefore, if one can show a lower bound for (/,;,) that matches the upper bound up to a constant, then by a simple sampling argument (on either an / bipartite graph or an / bipartite graph that achieves the maximum edge number), we can show that for all ,, one of the above two upper bounds is tight up to a constant.
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Once such a sequence is found, a similar question can be asked with f(n) taking the role of 1/n, and so on. In this way it is possible to investigate the borderline between divergence and convergence of infinite series. Using the integral test for convergence, one can show (see below) that, for every natural number k, the series