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Board
/sci/ - Science & Math Why do people deny this simple fact? You can prove it in like 4 lines.
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>>13886683
>You can prove it in like 4 lines.
no you can't, you can write down a pseudo-proof (emphasis on "pseud") in 4 lines
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>>13886683
the "proof" always involves some voodoo that assigns a non-convergent series a finite sum. garbage in garbage out.
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>>13886683
infinities don't exist bucko
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S = 1 + 2 + 3 + 4 + ... =
1 + 3 + 5 + 7 + .... +
2 + 4 + 6 + 8 + .....= 1 + 3 + 5 + 7 + .. + 2S
Therefore 1 + 3 + 5 + 7 + ... = S - 2S = -S = 1/12. But it's also equal to
(1 + 3)+ (5 + 7) + .. = 4 + 12 + 20 + ... = 4(1 + 3 + 5 + 7 + ... )= 4/12. Therefore 3/12 = 0.
QED
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>>13886683
anyone have the screenshot of the physics book deboonking the idea that this can be used in QM?
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>>13887105
youre not allowed to rearrange infinite sums, not even convergent ones
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OP = Faggot
I proved it in 1.
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>>13886683
If its the "proof" from numberphile video consider killing yourself.
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>>13886683
This result stems from the analytic continuation of the Riemann zeta function in the reals below 1. What you have written is retarded.
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>>13887155
You can the absolutely convergent sums.
$\sum_{n=1}^{\inf} \lvert a_n \rvert$
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>>13887319
Retard, I proved it in -1.
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>>13886694
>>13887088
>>13887328
You can prove it with a simple exchange of limits.
https://math.stackexchange.com/questions/1327812/limit-approach-to-finding-1234-ldots
\begin{align} \sum_{n=1}^\infty n &= \sum_{n=1}^\infty n \lim_{\varepsilon \to 0^+} \mathrm{e}^{-n \varepsilon} \left(1 - \frac{n \varepsilon}{2} \right) \\ &\stackrel{\star}{=} \lim_{\varepsilon \to 0^+} \sum_{n=1}^\infty n \mathrm{e}^{-n \varepsilon} \left(1 - \frac{n \varepsilon}{2} \right) \\ &= \lim_{\varepsilon \to 0^+} -\frac{\mathrm{e}^\varepsilon (\mathrm{e}^\varepsilon \varepsilon + \varepsilon - 2 \mathrm{e}^\varepsilon + 2)}{2 (\mathrm{e}^\varepsilon - 1)^3} \\ &= -\frac{1}{12} \end{align}
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popmath youtube shitting out another armchair complex analyst
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>>13886683
there is no such thing as an infinite sum. exoteric mathematics is fucking retarded
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>>13888649
Infinite sums are hardly esoteric
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>>13887986
>just exchange the limits bro
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>>13887986
Exchange of limits if only valid when both limits are convergent, you tard
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>>13886683
>Why do people deny this simple fact?
Because you keep using the word "sum" to describe your process, even though it violates the basic properties of summation.
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Various theories which perturb the theory of analysis a bit end up with OP's result, which I suppose is an interesting insight.
Anyway, in analysis

$\dfrac {1} { \log(1+x)} = \dfrac {1} {x} \frac {1} {1 - \left(\frac{1}{2}x - \frac{1}{3}x^2 + {\mathcal O}(x^3)\right) } = \dfrac {1} {x} + \dfrac{1}{2} - \dfrac{1}{12} x + {\mathcal O}(x^2)$

$\lim_{q \to 1} n\, q^n = n$

and so

$\lim_{q \to 1} \lim_{k \to \infty} \left( \sum_{n=0}^k n\, q^n - \int_0^k n\, q^n\, {\mathrm d}n \right) = - \frac{1}{12}$

In analysis,
$\lim_{q \to 1} \lim_{k \to \infty} \sum_{n=0}^k n\, q^n$
on its own does not exist.
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>>13886683
gr8 b8 m8 i r8 it 8/8
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>>13888959
>the real numbers aren't native enough to our consciousness we need 10 chapters to formalize the concept of greater than
>using a different number system? NONONONONO you can't do that
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>>13886683
People don't actually believe this, right? Why would it be anything but infinity?
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>>13888873
Not if you are willing to extend summation to divergent series, retard. That's the whole point.
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>>13888879
>it violates the basic properties of summation
Such as?
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>>13890004
Such as the sum being greater than any of the summands when the summands are positive.
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astounding
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>>13890013
And how do you prove that?
Induction doesn't work, since we're talking about infinite sums, not arbitrarily long but finite sums.
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>>13890071
>And how do you prove that?
Prove what? That the summands don't suddenly become negative past a certain point? Because I'm pretty sure every positive integer is greater than -1/12.
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>>13890119
He's making the point that arithmetic doesn't involve infinite numbers, so to prove that an infinite sum of positive number is positive needs heavy machinery (norms on the reals).
E.g. an infinite sum of rationals doesn't need to be rational either, e.g.

$\sum_{n=1}^N\dfrac{6}{n^2}$
is rational for all N, but

$\lim_{N\to \infty}\sum_{n=1}^N\dfrac{6}{n^2}=\pi^2$
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>>13890131
Are the summands in this case all positive? Is the sum negative? Is a negative number less than all positive numbers? Are you disputing that the sum of only positive numbers can never be smaller than the summands for finite or infinite convergent series?
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>>13887986
> he has not read Terrence Tao Analysis I and II
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>>13887088
The sum doesn't pass the term test, that is the lim of the term doesn't equal zero therefore it doesn't converge, right? Just making sure I understand your point.

So basically retards are manipulating a nonconvergent sum in an illegal way?
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>>13890269
There's always two steps when finding a limit:
1. Prove the limit exists
2. Find it's value, given that it exists
But there's nothing stopping you doing it in the opposite order, i.e.:
1. Assuming the limit exists, find the only value it can be
2. Prove that the series achieves that limit

Bullshit like -1/12 basically does the first part and just skips the second, then pretends it's some profound thing.
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$\sum_{n=0}^\infty (-1+\varepsilon)^n=\frac{1}{1-(-1+\varepsilon)}=\frac{1}{2-\varepsilon}$

So in the limit
"$\sum_{n=0}^\infty (-1)^n=\frac{1}{2}$"

Except in analysis we can't take the limit of a proposition, and the limit of on the right
"$\lim_{N\to\infty}\sum_{n=0}^N (-1)^n$"
does not exist in analysis.

The fact that many summation methods end up in
$\sum_{n=0}^\infty (-1)^n=\frac{1}{2}$
speaks for the fact that there's some algebraic aspects to this equation that are sensible, but as far as the classical theory of real numbers go, this doesn't hold. Either.
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1 + 2 + 3 + ... n equals a negative number. Totally not retarded at all.
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>>13890899
>1 + 2 + 3 + ... n equals a negative number.
No one says that.
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>>13887328
This is literally the only person has said anything at all intelligent on the subject. No, 1+2+3+... does not converge to -1/12 by the regular definition of convergence. People aren't actually saying that. They are just saying what this guy said

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