/sci/ - Science & Math » Thread #10873690

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Downtime Math #3

Anonymous Thu 08 Aug 14:56:16 2019 No.10873690 View Reply Original Report

Quoted By: >>10873701 >>10873720

Hello again /sci/
This one should be pretty easy.
Last post: >>10873196

Anonymous

Anonymous Thu 08 Aug 2019 15:01:25 No.10873701 Report

Quoted By: >>10873731

>>10873690
2019/(2018^2)?

Anonymous

Anonymous Thu 08 Aug 2019 15:02:44 No.10873703 Report

Quoted By:

$\sum\limits_{n=1}^{\infty} Anx^n = \frac{Ax}{(1-n)^2}$

$\sum\limits_{n=1}^{\infty} n(\frac{1}{2019})^n = \frac{\frac{1}{2019}}{(1-\frac{1}{2019})^2}$

Anonymous

Anonymous Thu 08 Aug 2019 15:03:46 No.10873705 Report

Quoted By: >>10873731 >>10873737

$\sum\limits_{n=1}^{\infty} Anx^n = \frac{Ax}{(1-x)^2}$

$\sum\limits_{n=1}^{\infty} n(\frac{1}{2019})^n = \frac{\frac{1}{2019}}{(1-\frac{1}{2019})^2}$

Anonymous

Anonymous Thu 08 Aug 2019 15:08:09 No.10873720 Report

Quoted By: >>10873731

>>10873690
You just use the closed form of the geometric series on an infinite sum of closed forms of other geometric series. ((1/2019)/(1-1/2019))/(1-1/2019) = (1/2019)/(1-1/2019)^2

Anonymous

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Anonymous Thu 08 Aug 2019 15:12:58 No.10873731 Report

Quoted By:

>>10873701
>>10873705
>>10873720
Far too easy I guess.

Anonymous

Anonymous Thu 08 Aug 2019 15:16:01 No.10873737 Report

Quoted By:

>>10873705
$S = 1 + x^2 + x^3 + \cdots = \frac{1}{1-x}$
$R = x + 2x^2 + 3x^3 + \cdots = x\frac{ds}{dx} = x\frac{d}{dx}[\frac{1}{1-x}] = \frac{x}{(1-x)^2}$

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