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

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Homework Statement

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

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

Homework Equations

lim n->\infty=\alpha _{n}

|\alpha-\alpha _{n} |\leq K*|\beta n|

The Attempt at a Solution

I found the limit of alpha \alpha _{n}= 2

Then,

|\frac{2*n^{2}+n+1}{n^{2}-3 -2}|=\frac{n+7}{|n^{2}-3|}

Here I'm stock.

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[morphism]

What do you mean by rate of convergence?

 

[CompuChip]

|\frac{2*n^{2}+n+1}{n^{2}-3 -2}|=\frac{n+7}{|n^{2}-3|}

Heh, you misplaced a bracket, obviously you meant
|\frac{2*n^{2}+n+1}{n^{2}-3} -2|=\frac{n+7}{|n^{2}-3|}

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
\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 }<br />
and work it out as you did above, then take the limit
\lim_{n \to \infty} \frac{a_{n+1} - 2}{a_n - 2}.

I don't know what definition you use though.

 

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Yes I misplaced a bracket, thanks compuchip.

Rate of convergence definition.

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

| \alpha _{n} - \alpha| \leq K|\beta _{n}|, for large n,

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

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