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The following image is from Geometric Series Proofs: An Annotated Bibliography.

enter image description here

Please explain why it is said that:

"$ON$ is the limit of the sum $1+x+\dots$."

Thank you.

Edit: I guess what through me off was the word "limit"!

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    $\begingroup$ If anyone is interested, several years ago I managed to trace this approach back to Milutin Milankovitch [Milankovitsch], Eine graphische darstellung der geometrischen progressionen, Zeitschrift für mathematischen und naturwissenschaftlichen 40 (1909), 329. I do not know whether this idea had been published earlier than 1909. In the 1950s Constantin P. Orloff published several follow-up papers on this in a Yugoslavian journal. $\endgroup$ – Dave L. Renfro Jun 6 '14 at 14:08
  • $\begingroup$ @DaveL.Renfro I appreciate the historical detail. $\endgroup$ – NoChance Jun 9 '14 at 20:38
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Notice that the horizontal lines of length $1,x,x^2,x^3,\ldots$ union to form a line of length $ON$.

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There are two ways to travel the horizontal distance from $O$ to $N$. The easiest way is to go via the line $ON$. Alternatively, you could follow the staircase, only counting the movement in the horizontal direction. The first stair has horizontal movement $1$, the second has horizontal movement $x$, and so on. However, once you've arrived at $N$, it shouldn't matter which way you went, the distances are the same. Therefore (the length of the line segment) $ON$ is the same as the limit of the sum $1 + x + x^2 + \dots$

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Why are the length of the line segments AB, CD, EF, etc, $x$, $x^2$, $x^3$, etc, respectively? I perceive that multiplying each base with the slope $x$ renders these results, but I do not understand how.

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Not really an answer to the question. If anyone objects I'll delete it.

I came across this trying to find an alternative expression for $\Delta$ in the picture. I thought it looked kinda nice. Similar to the picture above but not exactly the same. Geometric series

The length marked as $\Delta$ in the picture is of course simply: $1-\cos{(\beta-\phi)}$. Looking for an alternative way to calculate $\Delta$ we get the geometric series:

First approximate $\Delta$ as $\Delta_1$. Now we have to subtract the red part in the picture. Going over to the other side we see $\Delta_2$ contains the length of that red part but has an additional part (the green part in the picture). If we subtract that and repeat this process infinitely we must get $\Delta$.

This results in :

$1-\cos{(\beta-\phi)}=\sin^2{(\beta-\phi)}\left\{ 1-\cos{(\beta-\phi)} +\cos^2{(\beta-\phi)} -\cos^3{(\beta-\phi)} + \cdots \right\} \implies $

$ \frac{1-\cos{(\beta-\phi)}}{\sin^2{(\beta-\phi)}} = \frac{1-\cos{(\beta-\phi)}}{1-\cos^2{(\beta-\phi)}}= \frac{1-\cos{(\beta-\phi)}}{(1+\cos{(\beta-\phi)})(1-\cos{(\beta-\phi)})} = $

$ \frac{1}{1+\cos{(\beta-\phi)} }= 1-\cos{(\beta-\phi)} +\cos^2{(\beta-\phi)} -\cos^3{(\beta-\phi)} + \cdots $

The geometric series.

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  • $\begingroup$ Thanks for posting, I guess its too complex for me to get at the moment. $\endgroup$ – NoChance Feb 8 at 21:23
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    $\begingroup$ @NoChance I didn't completely work out the calculations. But it's fairly easy if you know where to look. I know it's a very old post, but if you're still interested let me know. In that case I'll work it out in a bit more detail. $\endgroup$ – Rutger Moody Feb 8 at 21:29
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    $\begingroup$ Thanks for willing to help, I will give it a try and may call on your help if you have time, cheers! $\endgroup$ – NoChance Feb 9 at 0:13

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