Series test for convergence/divergence

05112018, 06:56 PM
Post: #10




RE: Series test for convergence/divergence
sum(1/n^alpha,n) is convergent if alpha>1, divergent for alpha<=1 (this is easy to prove by comparing with int(1/x^alpha,x)). If f(n) is equivalent to g(n) and g(n) has constant sign then sum(f(n),n) and sum(g(n),n) are both convergent or both divergent.
It's more complicated for non constant sign term, like sum((1)^n/n^alpha,n) is convergent for alpha>0, but sum((1)^n/f(n),n) may be divergent even if f(n) is equivalent to n^alpha, for this kind of series you will usually need a series expansion of f(n) at n=inf until the remainder of the expansion is O(1/n^alpha) with alpha>1. Like for example sum((1)^n/sqrt(n+(1)^n*sqrt(n)),n) 

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Messages In This Thread 
Series test for convergence/divergence  DrD  05092018, 11:15 AM
RE: Series test for convergence/divergence  parisse  05092018, 12:33 PM
RE: Series test for convergence/divergence  DrD  05102018, 08:25 AM
RE: Series test for convergence/divergence  parisse  05102018, 01:31 PM
RE: Series test for convergence/divergence  DrD  05102018, 02:09 PM
RE: Series test for convergence/divergence  parisse  05112018, 03:30 PM
RE: Series test for convergence/divergence  DrD  05112018, 04:17 PM
RE: Series test for convergence/divergence  Benjer  05112018, 04:25 PM
RE: Series test for convergence/divergence  toshk  05112018, 06:28 PM
RE: Series test for convergence/divergence  parisse  05112018 06:56 PM
RE: Series test for convergence/divergence  Benjer  05112018, 11:17 PM

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