I've been given a homework question as follows:

Let $a$ be a positive real number, find a method to approximate $1/a$ without ever having to divide, with a sequence $x_k$ such that $x_k \to 1/a$. Explain why your solution works and and as much information as you can about the error. Hint. Think of a geometric series whose sum is $1/a$.

I've been going through all the series I can think of and my first question is, is there a way to create a geometric series whose sum is of your choice?

(I'm also finding it confusing how a sequence could converge to some number whilst the series also sums to that same number)

  • $\begingroup$ What does x sub k mean ? $\endgroup$ – Amr Feb 7 '13 at 13:13
  • $\begingroup$ and what does dividing by x sub k mean ? $\endgroup$ – Amr Feb 7 '13 at 13:15
  • $\begingroup$ @Jun OK. and waht does dividing by x_k mean ? $\endgroup$ – Amr Feb 7 '13 at 13:16
  • $\begingroup$ It means finding some method of approximation which does not include the use of division. $\endgroup$ – McT Feb 7 '13 at 13:20
  • $\begingroup$ Concerning your parenthetical comment: The sequence $(x_k)$ is the sequence of partial sums of the series. $\endgroup$ – David Mitra Feb 7 '13 at 13:24

Sum the series

$$S_k = \sum_{n=0}^{k-1} (1-a)^n = \frac{1-(1-a)^k}{1-(1-a)} = \frac{1}{a} [1-(1-a)^k]$$


$$\lim_{k \rightarrow \infty} S_k = \frac{1}{a}$$

This only works, of course, when $|1-a| < 1$.


Apply Newton's method to the equation $$ f(x)=\frac1x-a=0. $$ Choose a first approximation $x_0$ and define $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n(2-a\,x_n),\quad n\ge1. $$


$$\frac{1}{a}=\frac{1}{1-(1-a)}=\sum_{i=0}^{\infty}(1-a)^i$$ $$|1-a|<1\Rightarrow -1<1-a<1\Rightarrow 0<a<2,$$


The existing answers basically cover it, but I thought I would post an answer that works for all $a$. Choose $\varepsilon$ such that $0<\varepsilon a < 1$, and let $y_{k}=\varepsilon(1-\varepsilon a)^{k}$. Then $$ \sum_{k=0}^{\infty}y_k= \varepsilon\sum_{k=0}^{\infty}\left(1-\varepsilon a\right)^k=\frac{\varepsilon}{1-(1-\varepsilon a)}=\frac{1}{a} $$ and is convergent, so the partial sums $x_k\equiv\sum_{i=0}^{k}y_i \rightarrow 1/a$.


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