Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $

Please, help!

share|cite|improve this question
What does $S_n$, $A_n$ mean? – Hanul Jeon Feb 5 '13 at 8:47
up vote 2 down vote accepted

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 $k=0$.

share|cite|improve this answer
yes, it should be "-" and you solution is perfect – John Lennon Feb 5 '13 at 17:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.