# How to speed up the convergence?

I should you Aitken's method and the following formula :

$T_n = S_n - \dfrac{{A_{n+1}}^2}{A_{n+1}-A_{n+2}}$

in order to speed up $S_n=\sum_{k=1}^{n} (0.99)^k$

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What does $S_n$, $A_n$ mean? –  tetori Feb 5 '13 at 8:47

Okay even though this question isn't well posed, let me present an answer under the following assumptions.

You refear to the Aitken's method that is presented here in wikipedia. Dooes your method equal the following? $$Ax_n=x_n-\frac{(\Delta x_n)^2}{\Delta^2 x_n}$$ Where $\Delta x_{n}={(x_{n+1}-x_{n})},\$ and $\Delta^2 x_n=x_n -2x_{n+1} + x_{n+2}.$ At least, this is how it is written in wiki. If so, your $S_n$ corresponds to $x_n$ and your $A_n$ to $\Delta x_n$ and $T_n$ to $Ax_n$. Note that you have a $+$ between the two terms an wiki a $-$. Without any further information I don't know how to handle that, so I take the approach from wiki. $$\Delta x_n = x_{n+1} -x_n = \sum_{k=1}^{n+1}0.99^k-\sum_{k=1}^{n}0.99^k=0.99^{n+1}$$ Furthermore $$\Delta^2 x_n = x_{n+2} -x_{n+1} -x_{n+1}+x_n = \sum_{k=1}^{n+2}0.99^k-\sum_{k=1}^{n+1}0.99^k-\sum_{k=1}^{n+1}0.99^k+\sum_{k=1}^{n}0.99^k \\= 0.99^{n+2}-0.99^{n+1}=(0.99-1)0.99^{n+1}=-0.01\cdot 0.99^{n+1}$$ Therefore $$\frac{(\Delta x_n)^2}{\Delta^2 x_n}=\frac{0.99^{2n+2}}{-0.01\cdot0.99^{n+1}} = -100\cdot 0.99^{n+1}$$ And $$Ax_n=x_n-\frac{(\Delta x_n)^2}{\Delta^2 x_n}=\sum_{k=1}^{n}0.99^k+100\cdot 0.99^{n+1}$$ It holds $S_n=\sum_{k=1}^{\infty}=99$ by the convergence of geometric series. Now if we plug $n=1$ into $Ax_n$ we already get $Ax_1=99$ so we converge immediatly. But this is acutally no surprise. $$Ax_n=\sum_{k=1}^{n}0.99^k+100\cdot 0.99^{n+1}= \frac{1-0.99^{n+1}}{1-0.99}-0.99^0+ 100\cdot 0.99^{n+1} = 99$$ So we have constant values for $Ax_n$. Note that the $0.99^0$ is due to the fact the most geometric series start at $k=0$.

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yes, it should be "-" and you solution is perfect –  John Lennon Feb 5 '13 at 17:25