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Anonymous Mon 23 Sep 14:50:16 2019 No.10996405 View Reply Original Report

Quoted By: >>10996407 >>10996425 >>10996438

whenever I come back to this type of problems I stop thinking. I can't find any examples on this exact type of sums, it makes me feel retarded as fuck

Anonymous

Anonymous Mon 23 Sep 2019 14:52:54 No.10996407 Report

Quoted By: >>10996544

>>10996405
Hmm... What about WHAT THEY FUCKING ARE, WHICH IS INFINITE GEOMETRIC SERIES???
holy fuck
That one is 3/(1 - 1/8) = 24/7.

Anonymous

Anonymous Mon 23 Sep 2019 15:02:37 No.10996425 Report

Quoted By:

>>10996405
why is this written retardedly and not just
$sum_{n=1}^{\infty} \frac{3}{8^n}]/eqn]$

Anonymous

Anonymous Mon 23 Sep 2019 15:03:44 No.10996428 Report

Quoted By:

why is this written retardedly and not just
$\sum_{n=1}^{\infty} \frac{3}{8^n}$

Anonymous

Anonymous Mon 23 Sep 2019 15:07:06 No.10996438 Report

Quoted By:

>>10996405
This is not a problem, it's just an infinite sum.

Anonymous

Anonymous Mon 23 Sep 2019 15:59:48 No.10996544 Report

Quoted By:

>>10996407
wait shit i was wrong, it's 3/7

Anonymous

Anonymous Mon 23 Sep 2019 17:17:26 No.10996757 Report

Quoted By:

$\sum_{n=1}^{\infty} \frac{3}{8^n}$
$= 3\sum_{n=1}^{\infty} \frac{1}{8^n}$
$= 3\frac{1-1/8}{1-1/8}\sum_{n=1}^{\infty} \frac{1}{8^n}$
$= \frac{3}{1-1/8}\left(\sum_{n=1}^{\infty} \frac{1}{8^n} - \frac{1}{8}\sum_{n=1}^{\infty} \frac{1}{8^{n}} \right)$
$= \frac{3}{1-1/8}\left(\sum_{n=1}^{\infty} \frac{1}{8^n} - \sum_{n=1}^{\infty} \frac{1}{8^{n+1}} \right)$
$= \frac{3}{1-1/8}\left(\frac{1}{8} + \sum_{n=2}^{\infty} \frac{1}{8^n} - \sum_{n=2}^{\infty} \frac{1}{8^{n}} \right)$
$= \frac{1}{8}\frac{3}{1-1/8}$
$= \frac{3}{8-1}$
$= \frac{3}{7}$

Anonymous

Anonymous Mon 23 Sep 2019 17:31:54 No.10996793 Report

Quoted By: >>10996828

it's the integral of 3/(8^x) from 1 to infinity

Anonymous

Anonymous Mon 23 Sep 2019 17:45:07 No.10996828 Report

Quoted By:

>>10996793
no it's really not.

First of all, solving these sorts of sums is how the integral is derived in the first place.
Second of all, what you are describing will be a series of boxes of width 1 which are lined up with the curve.

So you are approximately right, but not right.
Your idea is used in statistics quite often when you are approximating.

$ln(n!) = ln(n) + ln(n-1) + ln(n-2) + ......$

this looks very close to $\int_{1}^{n}ln(x)dx$ but it is only an approximation.
The error is on the order of ln(n)

Anonymous

Anonymous Mon 23 Sep 2019 18:13:45 No.10996900 Report

Quoted By:

The sume obviously converges to a value S.

S= 3/8 +3/64+3/512+...
S/8= 3/64+3/512+...=S-3/8
3/8=S-S/8
7S/8=3/8
S=3/7

This is not a hard problem for a genius like me oWo.

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