Rate of Convergence of \alpha _{n}: Find Limit & K

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Homework Help Overview

The discussion centers around finding the rate of convergence of the sequence defined by \(\alpha_n = \frac{2n^2+n+1}{n^2-3}\) as \(n\) approaches infinity. Participants are exploring the limit of this sequence and the associated constant \(K\) in the context of convergence rates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to determine the limit of \(\alpha_n\) and are discussing the definition of rate of convergence. There are questions about the meaning of "rate of convergence" and suggestions to compute the limit of the ratio of differences from the limit.

Discussion Status

Some participants have provided clarifications regarding the notation and definitions involved in the problem. There is an ongoing exploration of the mathematical expressions and their implications for understanding convergence.

Contextual Notes

One participant notes a potential error in the expression used for the convergence analysis, indicating a need for careful attention to mathematical details. The definition of rate of convergence is also being discussed, referencing a specific source.

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Homework Statement



Find the Rate of Convergence of [tex]\alpha = \frac{2*n^{2}+n+1}{n^{2}-3}[/tex]

n=1,2,3,...,...

Homework Equations



[tex]lim n->\infty=\alpha _{n}[/tex]

[tex]|\alpha-\alpha _{n} |\leq K*|\beta n|[/tex]

The Attempt at a Solution



I found the limit of alpha [tex]\alpha _{n}= 2[/tex]

Then,

[tex]|\frac{2*n^{2}+n+1}{n^{2}-3 -2}|=\frac{n+7}{|n^{2}-3|}[/tex]

Here I'm stock.

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What do you mean by rate of convergence?
 
Link- said:
[tex]|\frac{2*n^{2}+n+1}{n^{2}-3 -2}|=\frac{n+7}{|n^{2}-3|}[/tex]
Heh, you misplaced a bracket, obviously you meant
[tex]|\frac{2*n^{2}+n+1}{n^{2}-3} -2|=\frac{n+7}{|n^{2}-3|}[/tex]

I also wonder what you mean by rate of convergence, but taking the "simple" definition on this Wikipedia page I think you want to start out by filling in
[tex]\frac{a_{n+1} - 2}{a_n - 2} <br /> = \frac{ \frac{2n^2+n+1}{n^2-3} - 2 }{ \frac{2*n^{2}+n+1}{n^{2}-3} - 2 }[/tex]
and work it out as you did above, then take the limit
[tex]\lim_{n \to \infty} \frac{a_{n+1} - 2}{a_n - 2}[/tex].

I don't know what definition you use though.
 
Yes I misplaced a bracket, thanks compuchip.

Rate of convergence definition.

Suppose [tex]\left \{\beta _{n} \right\}}^{\infty}_{n=1}[/tex] is a sequence known to converge to zero, and [tex]\left\{\alpha _{n} \right\} ^{\infty}_{n=1}[/tex] converges to a number [tex]\alpha[/tex]. If a positive constant K ecists with

[tex]| \alpha _{n} - \alpha| \leq K|\beta _{n}|[/tex], for large n,

then we way that [tex]\left\{\alpha _{n} \right\} ^{\infty}_{n=1}[/tex] converges to [tex]\alpha[/tex] with rate of convergence [tex]O( \beta _{n})[/tex]. It is idndicated by writing [tex]\alpha _{n}=\alpha + O( \beta _{n})[/tex].

Obtained from "Numerical Analysis 8th ed", by Burden and Faires.
 

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