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 $S_n=\sum_{k=1}^{n} (0.99)^k$

What does $S_n$, $A_n$ mean? – Hanul Jeon Feb 5 '13 at 8:47
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 $k=0$.