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.

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Hint: Think of the Riemann Zeta function.

- dumbcow

umm the infinite sum of the natural numbers is infinity

- anonymous

Nope, sorry. I was asking for the analytical sum.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- dumbcow

sum of an arithmetic series
\[S_n = \frac{n(a_1 +a_n)}{2}\]
for set of natural numbers
\[S_n = \lim_{n \rightarrow \infty}\frac{n(n-1)}{2}\]
which obviously goes to infinity
i guess im not understanding what you're looking for

- anonymous

Yes, the sum diverges.
The question is not such, I'm asking for an analytical, total sum of the set of natural numbers.

- dumbcow

hmm i guess i never learned this or remember it...analyzing the cardinality of the sum in terms of infinity? or something like that

- dumbcow

my math knowledge stops at number theory and advanced analysis :|

- anonymous

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 \(\mathbb{N}\))

- ParthKohli

\[{n \over 2}\left(a_1 + a_n \right)\]So here\[{\lim_{n \to \infty} {n \over 2}(1 + a_n)}\]Though, the last natural number can NOT be determined.

- anonymous

Nope, it's an actual discreet value. Again, you can't use the \(\frac{n(n+1)}{2}\), because those are partial sums.

- anonymous

*

- anonymous

Hmm?

- anonymous

lol...i mean i would like to know what is the answer :)

- anonymous

Hint 2: The series is not Abel-Summable, but can be manipulated to do such, and manipulated back.

- anonymous

@mukushla Oh, all right, haha, if no one gets it by tomorrow I'll post up the answer.

- mathslover

Hint 3 : try to search on google ... if no then follow mukushla's spirit :P

- anonymous

(Although, I AM surprised no one has gotten this... how many of you have taken functional analysis/analytic class numbers?)

- mathslover

not yet sorry

- Compassionate

*Watches you guys debate while eating cereal.*
Lolgoodguessbrobutstillwrong.jpg

- Compassionate

This is boring. Prove it as N approaches infinity and put it in a summation, then use the formula stated above. Toodles.

- anonymous

@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.

- Compassionate

RolloverlolnobackspacedeletethatCTRLALDELnopewrong.avi

- anonymous

That would be for whose post...?

- anonymous

All right, well, here's the answer. It might seem a little "out-there" for the first-time 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 \(S(x)\):
\[\begin{align}
S(x)&=1+x+x^2+\cdots\implies\\
xS(x)&=x+x^2+x^3+\cdots\implies\\
1+xS(x)&=1+x+x^2+\cdots=S(x)\implies\\
S(x)&=\frac{1}{1-x}=1+x+x^2+\cdots
\end{align}
\]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:
\[
S'(x)=\frac{1}{(x-1)^2}=0+1+2x+3x^2\cdots
\]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):
\[
S'(-1)=\frac{1}{4}=1-2+3-4+\cdots
\](This was the second hint I gave, about Abel-summability. 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:
\[
\begin{align}
\zeta(s)&=1^{-s}+2^{-s}+3^{-s}+\cdots\implies\\
(1-2^{-s})\zeta(s)&=1^{-s}+2^{-s}-2^{-s}+3^{-s}+\cdots=1^{-s}+0+3^{-s}+\cdots\implies\\
(1-4^{-s})\zeta(s)&=1^{-s}-2^{-s}+3^{-s}+\cdots
\end{align}
\]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:
\[\begin{align}
(1-4^{1})\zeta(-1)&=1-2+3-4+\cdots=\frac{1}{4}\\
(-3)\zeta(-1)&=\frac{1}{4}\\
\zeta(-1)&=-\frac{1}{12}
\end{align}
\]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

- anonymous

Oh, also:
http://www.wolframalpha.com/input/?i=zeta%28-1%29

- anonymous

Well, the answer's up:
@dumbcow, @mukushla, @mathslover, @ParthKohli, @Compassionate

- anonymous

Oh, sorry, I had a slight error on one part, but it's an easy correction and still gives the same result, change
\[
(1-4^{-s})\text{ to }(1-2\cdot2^{-s})
\]

- dumbcow

yes never took these classes...but you are saying the sum of all positive integers equals a negative fraction huh

- anonymous

I know, right? I said the same thing, haha, it's kinda ridiculous, but it's true.

- anonymous

o.O

Looking for something else?

Not the answer you are looking for? Search for more explanations.