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If p ≤ 0, then the nth-term test identifies the series as divergent. If 0 < p ≤ 1, then the nth-term test is inconclusive, but the series is divergent by the integral test for convergence. If 1 < p, then the nth-term test is inconclusive, but the series is convergent by the integral test for convergence.
Leibniz integral rule; Definitions; Antiderivative; Integral ... This is also known as the nth-term test, test for divergence, or the divergence test.
The n-Fibonacci constant is the ratio toward which adjacent -Fibonacci numbers tend; it is also called the n th metallic mean, and it is the only positive root of =. For example, the case of n = 1 {\displaystyle n=1} is 1 + 5 2 {\displaystyle {\frac {1+{\sqrt {5}}}{2}}} , or the golden ratio , and the case of n = 2 {\displaystyle n=2} is 1 + 2 ...
The figure shows that 8 can be decomposed into 5 (the number of ways to climb 4 steps, followed by a single-step) plus 3 (the number of ways to climb 3 steps, followed by a double-step). The same reasoning is applied recursively until a single step, of which there is only one way to climb.
In mathematics, a recurrence relation is an equation according to which the th term of a sequence of numbers is equal to some combination of the previous terms. Often, only k {\displaystyle k} previous terms of the sequence appear in the equation, for a parameter k {\displaystyle k} that is independent of n {\displaystyle n} ; this number k ...
It is assumed that the value of a function f defined on [,] is known at + equally spaced points: < < <.There are two classes of Newton–Cotes quadrature: they are called "closed" when = and =, i.e. they use the function values at the interval endpoints, and "open" when > and <, i.e. they do not use the function values at the endpoints.
Combining two consecutive steps of these methods into a single test, one gets a rate of convergence of 9, at the cost of 6 polynomial evaluations (with Horner's rule). On the other hand, combining three steps of Newtons method gives a rate of convergence of 8 at the cost of the same number of polynomial evaluation.
The nth partial sum of the series is the triangular number ∑ k = 1 n k = n ( n + 1 ) 2 , {\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},} which increases without bound as n goes to infinity .