# How to determinate ending point using series?

Consider that I have a path which start at point (0,0).

I need to find the ending point of the path using series.

The path look like this :

Any idea on how to start that ?

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The $x$ coordinate is $$8 \left(1 - \frac{3}{4} + \frac{3^2}{4^2} - \frac{3^3}{4^3} + \ldots \right)$$ and the $y$ coordinate is $$6 \left(1 - \frac{4}{5} + \frac{4^2}{5^2} - \frac{4^3}{5^3} + \ldots \right).$$ So you have to find the value of two geometric series.

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Alright, but could you please explain how you found they were the coordinate as this is mainly what I can't figure out? – MathLearner Jun 8 '14 at 7:34
@ProgrammerJeff The vertical parts do not contribute to the $x$ coordinate. So the $x$ coordinate is $8 - 6 + 9/2 - \ldots$. Similarly the horizontal parts do not contribute to the $y$ coordinate. – WimC Jun 8 '14 at 7:37
Not too sure to understand.. The answer would be $(\frac{32}{7},\frac{30}{9})$ ? – MathLearner Jun 8 '14 at 7:46
Even tho you did not answer, I tried another way and arrived to the same answer so you did well explain it I was just unconfident. Anyway, thanks. – MathLearner Jun 9 '14 at 23:18

The horizontal line co-ordinates are

$$8 , 6, \frac{9}{2}, \frac{27}{8}, ???$$

These values give away the pattern,

$$\frac{9}{2}, \frac{27}{8}$$

WimC has explained the formula in his answer above, but even from this, I can deduce that the preceding values should be

$$\frac{1}{0.125}, \frac{3}{0.5}, \frac{9}{2}, \frac{27}{8}$$

This would indeed keep with the first 2 horizontal values in your example

$$\frac{1}{0.125} == 8, \frac{3}{0.5} == 6$$

If I apply the same pattern to the inner part of the series, the inner most horizontal ... values should be.

$$8, 6, \frac{9}{2}, \frac{27}{8}, \frac{81}{32}, \frac{243}{256}$$

You can apply the same approach for the vertical value as well as they follow a similar pattern.

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