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Numbers and Elementary Mathematics
Published in Dan Zwillinger, CRC Standard Mathematical Tables and Formulas, 2018
If |anan+1|=1+pn+Annq $ |\frac{{a_{n} }}{{a_{n + 1} }}| = 1 + \frac{p}{n} + \frac{{A_{n} }}{{n^{q} }} $ , for sufficiently large n, where q > 1 and the sequence {An} is bounded, mpote nte series is absolutely convergent if and only ua p > 1.Alternating series test
Solutions and Answers to Selected Exercises
Published in John Srdjan Petrovic, Advanced Calculus, 2020
so the radius of convergence is R = 1. Further, when x = 1 we have ∑n=0∞(−1)n3n+1 which converges by Alternating Series Test. Thus, the desired result is f(1).
Computational and numerical developments
Published in Seán M. Stewart, R. Barry Johnson, Blackbody Radiation, 2016
Seán M. Stewart, R. Barry Johnson
If the series given by Eq. (3.66) is used to estimate the value for the general fractional function by truncating it after N terms, the size of the error can be readily estimated since the series alternates as B2k alternates between positive and negative values. An estimate for the size of the error is given by the Alternating Series Estimation Theorem which states that if an alternating series is truncating after summing the first N terms the error will be no larger in magnitude than the size of the first neglected term in the sum.
Sum of alternating-like series as definite integrals
Published in International Journal of Mathematical Education in Science and Technology, 2022
The goal of the present note is to contribute to this point by summing the family of series where the sequence is k-periodic, i.e. , for all integers and satisfies We usually use two well-known methods. The first method, to establish the convergence of (1) is Dirichlet's test (Rudin, 1976, p. 70): if is a decreasing sequence of real numbers such that as , and a sequence of complex numbers with bounded partial sums, then converges. The case corresponds to Leibniz's test (also called alternating series test). Our second tool is Abel's continuity theorem for power series (Rudin, 1976, p. 174): if the series of complex numbers converges, then the power series converges for and We will prove that this limit reduces the sum of (1) to the definite integral which is finite since the factor 1−t can be eliminated from the numerator and denominator of fraction, thanks to the condition (2).