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LolWolf
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This shouldn't be too hard (for the peeps, here, at least):
Give an analytical, definite value for the sum of the entire set \(\mathbb{N}\) starting from its lowest element and transversing through it in order.
Then, prove it.
 one year ago
 one year ago
LolWolf Group Title
This shouldn't be too hard (for the peeps, here, at least): Give an analytical, definite value for the sum of the entire set \(\mathbb{N}\) starting from its lowest element and transversing through it in order. Then, prove it.
 one year ago
 one year ago

This Question is Closed

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Hint: Think of the Riemann Zeta function.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
umm the infinite sum of the natural numbers is infinity
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Nope, sorry. I was asking for the analytical sum.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
sum of an arithmetic series \[S_n = \frac{n(a_1 +a_n)}{2}\] for set of natural numbers \[S_n = \lim_{n \rightarrow \infty}\frac{n(n1)}{2}\] which obviously goes to infinity i guess im not understanding what you're looking for
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yes, the sum diverges. The question is not such, I'm asking for an analytical, total sum of the set of natural numbers.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
hmm i guess i never learned this or remember it...analyzing the cardinality of the sum in terms of infinity? or something like that
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
my math knowledge stops at number theory and advanced analysis :
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Remember that when you use the method of taking \(n\to\infty\) and \(\frac{n(n+1)}{2}\) you are receiving a partial sum. Such will always diverge, and thus end up at infinity. And, nah, it's not morphisms for countability (i.e. \(\mathbb{Q^+}\) is countable by \(\mathbb{N}\))
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[{n \over 2}\left(a_1 + a_n \right)\]So here\[{\lim_{n \to \infty} {n \over 2}(1 + a_n)}\]Though, the last natural number can NOT be determined.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Nope, it's an actual discreet value. Again, you can't use the \(\frac{n(n+1)}{2}\), because those are partial sums.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
lol...i mean i would like to know what is the answer :)
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Hint 2: The series is not AbelSummable, but can be manipulated to do such, and manipulated back.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
@mukushla Oh, all right, haha, if no one gets it by tomorrow I'll post up the answer.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Hint 3 : try to search on google ... if no then follow mukushla's spirit :P
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
(Although, I AM surprised no one has gotten this... how many of you have taken functional analysis/analytic class numbers?)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
not yet sorry
 one year ago

Compassionate Group TitleBest ResponseYou've already chosen the best response.0
*Watches you guys debate while eating cereal.* Lolgoodguessbrobutstillwrong.jpg
 one year ago

Compassionate Group TitleBest ResponseYou've already chosen the best response.0
This is boring. Prove it as N approaches infinity and put it in a summation, then use the formula stated above. Toodles.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
@Compassionate Lol, if it was that easy, don't you think that people would have gotten it within the first few minutes? Try doing that, yourself, see if the result is finite.
 one year ago

Compassionate Group TitleBest ResponseYou've already chosen the best response.0
RolloverlolnobackspacedeletethatCTRLALDELnopewrong.avi
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
That would be for whose post...?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
All right, well, here's the answer. It might seem a little "outthere" for the firsttime reader (I would say it would even look almost magical), but it's not that imaginative, once you're used to these manipulations: Starting from the geometric sum \(S(x)\): \[\begin{align} S(x)&=1+x+x^2+\cdots\implies\\ xS(x)&=x+x^2+x^3+\cdots\implies\\ 1+xS(x)&=1+x+x^2+\cdots=S(x)\implies\\ S(x)&=\frac{1}{1x}=1+x+x^2+\cdots \end{align} \]All right, so, we take the derivative (this is one of those apparently 'magical' parts, which becomes pretty mundane after a while). If you've studied generating functions, chances are this is nothing new: \[ S'(x)=\frac{1}{(x1)^2}=0+1+2x+3x^2\cdots \]But, we can't get a definite sum for this series, since \(S'(x)\) has a pole at \(x=1\). So, with some foresight, we plug in \(x=1\) (it will become obvious as of why, later): \[ S'(1)=\frac{1}{4}=12+34+\cdots \](This was the second hint I gave, about Abelsummability. This shouldn't have to be derived, I'm just doing it as a little extra). All right, so, we now use the function \(\zeta(s)\) as a manipulation aid: \[ \zeta(s)=\sum_{n=1}^\infty n^{s} \]We wish to negate every even term of the series, to be able to solve back for our own, final value, so we begin: \[ \begin{align} \zeta(s)&=1^{s}+2^{s}+3^{s}+\cdots\implies\\ (12^{s})\zeta(s)&=1^{s}+2^{s}2^{s}+3^{s}+\cdots=1^{s}+0+3^{s}+\cdots\implies\\ (14^{s})\zeta(s)&=1^{s}2^{s}+3^{s}+\cdots \end{align} \]Which is what we wished to find. Now, we know the value of this last series manipulation at \(s=1\), so we use this to find our value for \(\zeta(1)\), which is the sum of the natural set: \[\begin{align} (14^{1})\zeta(1)&=12+34+\cdots=\frac{1}{4}\\ (3)\zeta(1)&=\frac{1}{4}\\ \zeta(1)&=\frac{1}{12} \end{align} \]So... guess what? The sum of all natural number is \(\frac{1}{12}\). Enjoy! http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7_%C2%B7_%C2%B7
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Oh, also: http://www.wolframalpha.com/input/?i=zeta%281%29
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, the answer's up: @dumbcow, @mukushla, @mathslover, @ParthKohli, @Compassionate
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Oh, sorry, I had a slight error on one part, but it's an easy correction and still gives the same result, change \[ (14^{s})\text{ to }(12\cdot2^{s}) \]
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
yes never took these classes...but you are saying the sum of all positive integers equals a negative fraction huh
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
I know, right? I said the same thing, haha, it's kinda ridiculous, but it's true.
 one year ago
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