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Take a look at this page: http://wiki.ubc.ca/Keynesian_Multiplier
Why can you find out the sum of the geometric series just by dividing the mps by 1?

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I see no MPS, only MPC. Is that what you mean? And what is divided by 1? I can't see where that is. Can you be a little bit more specific? –  hejseb Nov 19 '13 at 16:41
ah sorry lol, MPS equals to 1 - MPC –  Mouse Hello Nov 19 '13 at 16:49
@hejseb is there any other thing ambiguous in my question? –  Mouse Hello Nov 19 '13 at 16:56
Oh, I see. I should probably have figured that out myself... See my answer below. Hope it clears some confusion up. –  hejseb Nov 19 '13 at 16:59

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Consider the series $$ 1+r+r^2+r^3+\cdots=\sum_{k=0}^\infty r^k $$ If $|r|<1$, then this is a geometric series whose sum is the well-known $$ \sum_{k=0}^\infty r^k=\frac{1}{1-r}. $$ So if you have $MPC$ instead of $r$ and $MPS=1-MPC$, then this is $$ \sum_{k=0}^\infty MPC^k=\frac{1}{1-MPC}=\frac{1}{MPS}. $$

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So, it's just a well known geometric series then? What is the geometric series called? Ah, and would you mind giving me a link to more of these well known geometric series? –  Mouse Hello Nov 19 '13 at 17:09
@MouseHello Have a look here: en.wikipedia.org/wiki/Geometric_series#Formula where you have this result and also for a finite number of terms. It's not called anything in particular - it's simply called a geometric series. –  hejseb Nov 19 '13 at 17:12
Cool, thanks :) –  Mouse Hello Nov 19 '13 at 17:15

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