How Can We Determine the Optimum Number of Terms for a Convergent Series?

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The discussion centers on determining the optimum number of terms, denoted as K, for a convergent series defined by the recurrence relation F(n, a_{n+2}, a_{n+1}, a_{n}) = n. The participants clarify that K represents the number of terms needed for convergence, while addressing the ambiguity of the term "optimum." A brute force algorithm is suggested for calculating K by evaluating a large number of terms, although the lack of a clear definition for "optimum" complicates the inquiry. The conversation emphasizes that a finite sum of terms will always converge, contrasting with the divergence of an infinite series.

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lokofer
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Let be the series...[tex]S=\sum_{n=0}^{\infty} a(n)[/tex]

where a(0)=1=a(1) and the rest of coefficients satisfy a recurrence relation (linear or non-linear) so [tex]F(n,a_{n+2} , a_{n+1},a_{n})=n[/tex] :-p :-p ..then my question is let's suppose that the series has an "optimum number of terms" K so if you take k-terms the series converges to a optimum value, otherwise the series (taking all terms) diverges) my question is how would we obtain this k and the sum of the series... a "brute force" algorithm would say that you take a big number of terms and solve the recurrence by using a computer...:rolleyes: :rolleyes:
 
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Since you have not defined optimum the question is impossible to answer.

When you use K and k are we supposed to think they refer to the same thing? What is a k-term? Do you just mean sum the first k terms (so why use the word convergent for a finite sum?), or do you mean to pick some infinite subset of the terms?

Your recurrence relation could very well be easy to solve (it is only a second order recurrence relation, as written.
 
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The sum of a finite number of finite terms always converges. If your ininite sum is divergent, then any stopping point will put you in the situation you mention.
 

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