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

 >> Anonymous Sun Nov 10 17:49:24 2019 No.11134779 File: 126 KB, 1131x622, 1549390961477.jpg [View same] [iqdb] [saucenao] [google] [report] >It took him a while, but Timmy finally realized that all math above basic addition is a huge meme
 >> Anonymous Sun Nov 10 17:54:58 2019 No.11134796   File: 1.03 MB, 3001x1836, 1497969901262.jpg [View same] [iqdb] [saucenao] [google] [report] Note that $\log(\frac{1}{1-X}) = \sum_n X^n/n [math]so the statement amounts to saying that the log has no finite limit.The log grows slowly, but indeed it grows forever, so it diverges.Besides, the sum even diverges if you just add up the reciprocals of the primes1/2 + 1/3 + 1/5 + 1/7 + ...  >> Anonymous Sun Nov 10 17:56:26 2019 No.11134798 File: 133 KB, 438x587, Dreyfuz.jpg [View same] [iqdb] [saucenao] [google] [report] Note that[math] \log(\frac{1}{1-X}) = \sum_n X^n/n$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 primes1/2 + 1/3 + 1/5 + 1/7 + ...
 >> Anonymous Sun Nov 10 18:00:27 2019 No.11134810 File: 16 KB, 600x315, q5OL30E.jpg [View same] [iqdb] [saucenao] [google] [report] >>11134796>infinite jest
 >> Anonymous Sun Nov 10 18:10:47 2019 No.11134839 >>11134779the fact that people will still fall for this bait pic really shows how underage this board is
 >> Anonymous Sun Nov 10 18:18:38 2019 No.11134865 >>11134839>tfw can't \mathbb{C}
 >> Anonymous Sun Nov 10 18:41:45 2019 No.11134908 >>11134779a^2+b^2=c^2(1)^2+(i)^2=c^21+(-1)=c^20=c^2c=0correct but that's not how the complex phasor plane operates
 >> Anonymous Sun Nov 10 18:51:01 2019 No.11134930 >>11134767If 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)$\sum_{k=1}^n \frac{1}{k} \approx \int_1^n \frac{dx}{x}=\log n$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
 >> Anonymous Sun Nov 10 19:01:53 2019 No.11134950 File: 167 KB, 500x737, 2cc.png [View same] [iqdb] [saucenao] [google] [report] >>11134930>>11134798so what is the slowest growing divergent series?
 >> Anonymous Sun Nov 10 19:05:26 2019 No.11134956 >>11134950I'm not sure if there is such a thing
 >> Anonymous Sun Nov 10 19:07:30 2019 No.11134964 >>11134950Suppose 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 $\ln\ln n$ if i recall correctly
 >> Anonymous Sun Nov 10 19:10:55 2019 No.11134972 >>11134767By the integral test it divergeslim x-->inf log(x)=infQED You're retarded
 >> Anonymous Sun Nov 10 19:15:38 2019 No.11134984 >>11134964>this series grows like lnlnnlnlnn if i recall correctlyhow do i calculate the growth of a given series
 >> Justin Ramsey Sun Nov 10 19:19:21 2019 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.
 >> Anonymous Sun Nov 10 19:21:43 2019 No.11135001 >>11134984If 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_primesOther ways include taylor series, fourier series, numerical estimates, etc, it's not an easy question. I'm no expert.>>11134993based schizo
 >> Anonymous Sun Nov 10 20:44:07 2019 No.11135190 >>11134779>t. luddite-troll
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