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11134767 No.11134767 [Reply] [Original] [archived.moe]

this is a meme right. this can’t possibly be correct

>> No.11134779
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11134779

>It took him a while, but Timmy finally realized that all math above basic addition is a huge meme

>> No.11134796 [DELETED] 
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11134796

Note that

[math] \log(\frac{1}{1-X}) = \sum_n X^n/n [math]

so the statement amounts to saying that the log has no finite limit.
The log grows slowly, but indeed it grows forever, so it diverges.

Besides, the sum even diverges if you just add up the reciprocals of the primes

1/2 + 1/3 + 1/5 + 1/7 + ...

>> No.11134798
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11134798

Note that

[math] \log(\frac{1}{1-X}) = \sum_n X^n/n [/math]

so the statement amounts to saying that the log has no finite limit.
The log grows slowly, but indeed it grows forever, so it diverges.

Besides, the sum even diverges if you just add up the reciprocals of the primes

1/2 + 1/3 + 1/5 + 1/7 + ...

>> No.11134810
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11134810

>>11134796
>infinite jest

>> No.11134839

>>11134779
the fact that people will still fall for this bait pic really shows how underage this board is

>> No.11134865

>>11134839
>tfw can't \mathbb{C}

>> No.11134908

>>11134779
a^2+b^2=c^2
(1)^2+(i)^2=c^2
1+(-1)=c^2
0=c^2
c=0

correct but that's not how the complex phasor plane operates

>> No.11134930

>>11134767
If you know integration, this is like a discrete approximation of the integral that defines the log function. (the other anon got the log behavior a slightly different way)

[math]\sum_{k=1}^n \frac{1}{k} \approx \int_1^n \frac{dx}{x}=\log n[/math]

The difference between the two sides in the approximation is the Euler gamma constant (in the limit n goes to infinity)

I agree it's pretty hard to conceive how slowly a log function increases, or on the flip side how fast an exponential increases

>> No.11134950
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11134950

>>11134930
>>11134798
so what is the slowest growing divergent series?

>> No.11134956

>>11134950
I'm not sure if there is such a thing

>> No.11134964

>>11134950
Suppose there's a slowest growing divergent series
[eqn]\sum_{n=0}^{\infty}a_n[/eqn], then
[eqn]\sum_{n=0}^{\infty}\frac{a_n}{2}[/eqn] grows slower and also diverges, therefore there is no slowest growing divergent series, since we can always find a slower one. However, we can find some really fucking slow ones, take [eqn]\sum_{p \ \text{prime}} \frac{1}{p}[/eqn] this series grows like [math]\ln\ln n[/math] if i recall correctly

>> No.11134972

>>11134767
By the integral test it diverges

lim x-->inf log(x)=inf

QED You're retarded

>> No.11134984

>>11134964
>this series grows like lnlnnlnlnn if i recall correctly
how do i calculate the growth of a given series

>> No.11134993

Sentences go on virtually forever because reality itself cannot intervene if it was personal against power or happened in proximity. I cannot stress deaths happen literally forever. Over and over again.

>> No.11135001

>>11134984
If you can find a closed form for the partial sum, you're done, it grows like that. If you can do an integral test sometimes you can compare it to that. That's where we get the log growth for the harmonic series, there are other ways to prove it too. Here's a couple ways to prove the log log growth of the reciprocal prime series:

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

Other ways include taylor series, fourier series, numerical estimates, etc, it's not an easy question. I'm no expert.
>>11134993
based schizo

>> No.11135190

>>11134779
>t. luddite-troll

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