Does 1+1+1+... = 1/2 or -1/2?

>1+1+1+...
Not a valid notation.

But you still understood it.

>Not a valid notation.
elaborate, why not?

1+1+1...=1+1+1...
a=1+1+1+...
0.999a=0.999+0.999+0.999...
a=0.999+0.999+0.999.../0.999
1+1+1+...=0.999(1+1+1+...)/0.999(1+1-1+1-...)
1+1+1+...=0.999/0.999
1+1+1+...=1
i crack the jee with this looool

>0.999+0.999+0.999.../0.999 = 0.999(1+1+1+...)/0.999(1+1-1+1-...)

I guess Veeky Forums is just brainlets.

yes

>1+1+1+... =
>...
what do you mean by ...?

I think you can guess.

Being a geometric series whose R is equal to or greater than one, it diverges and has no finite sum

Meme degreers will disagree

it means "etc"

unless he misunderstood it. hard to sayyyyyy

Brainlet pedants btfo.
terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

[math] \sum_{n=1}^\infty n^m z^n + (-1)^m\dfrac{m!}{\log(z)^{m+1}} = -\dfrac{1}{m+1} B_{m+1} + {\mathcal O}((z-1)) [/math]

So

[math] \lim_{z\to 1} \left( \sum_{n=1}^\infty n^m z^n + (-1)^m\dfrac{m!}{\log(z)^{m+1}} \right) = -\dfrac{1}{m+1} B_{m+1} [/math]

and e.g.

[math] \lim_{z\to 1} \left( \sum_{n=1}^\infty n z^n - \dfrac{1} {\log(z)^{2}} \right) = -\dfrac{1}{2} \dfrac{1}{6} [/math]

and

[math] \lim_{z\to 1} \left( \sum_{n=1}^\infty z^n + \dfrac{1}{\log(z)} \right) = - \dfrac{1}{2} [/math]

If you think it's [math] + \dfrac{1}{2} [/math], you may have been looking at a source which uses the alternative Bernoulli numbers (pic related), but those aren't the ones matching up with the zeta function, e.g. in formulas such as

[math] \zeta(2n) = \dfrac{(2\pi)^{2n}}{(2n)!} (-1)^{n+1} \dfrac{1}{2} B_{2n} [/math]

>those aren't the ones matching up with the zeta function
How do you know?

Depends on what you expect as an answer. You can take Mathematica.
I can also rant about
en.wikipedia.org/wiki/Euler–Maclaurin_formula#The_formula
where the standard BernoulliB's are used

ad.:
I remembered I also at one point tried to write down a derivation that only assumes a Taylor expansion, pic related
Although of course this avoids the Bernoullis

ad 2.:
okay I played around with this series regulation a bit and I find that OPs sum is indeed the one where several expressions that have 1+1+1+1+ as limit all are regularized by the same term (the -1/log(z)) and that one it gives you any result.

Specifically limit z->1 via

1 + z + z^2 + z^3 + z^4 ---> 1+1+1+1+

gives you +1/2 and limit z->1 via

z + z^2 + z^3 + z^4 ---> 1+1+1+1+

gives you -1/2.

ty

>series aren't valid

>the power of american education

[eqn]\sum_{n=0}^\infty z = \frac{1}{1-z} [/eqn]

now take the limit as z->1

[eqn]1+1+1... = \lim_{z \rightarrow 1} \frac{1}{1-z} = -\frac{1}{2}[/eqn]

viola!

Are you saying that the value of this series is someone related to wave/particle duality? If so why can't we just do experiments in quantum mechanics and verify if this sequence is correct, and therefore find the truth about the Riemann hypothesis?

he was trolling

nice limit of 1/(1-z)