If 1-1+1-1+1-1-1......=1/2, does sin(x) as x approaches infinity equal 0?
thats the sum of zeta functions you fucking retard
>>7944406
read it wrong, thought it was the -1/12 meme
But you are still a retard
>>7944409
Grandis series, why are you being so aggressive dude?
>>7944402
1/2 is the limit of the average of the partial sums
>>7944402
the sum does not converge absolutely therefore for every a where -inf<=a<=+inf there is a rearrangement of the partial sums that converges to a (Rudin 3.54).
>>7944659
You misinterpreted the theorem mate.
It's absolutely essential to the proof Rudin provides that the series converges conditionally but not absolutely.
It is not, in general, possible to rearrange a divergent series to be any extended real number.
>>7944402
>1-1+1-1+1-1-1......=1/2
It doesn't converge in the typical infinite sum fashion
>does sin(x) as x approaches infinity equal 0
No, and this is directly from the definition of a limit
It does if you use Cesaro convergence, but not if you use standard convergence, which is the only relevant mode of convergence.
For what it's worth, with [math]|z|<1[/math] you have
[math]\sum_{n=0}^\infty z^n = \dfrac{1}{1-z} [/math],
so that for a substitution [math]z = -1 + \varepsilon [/math] and any tiny [math]\varepsilon \in (0,1)[/math] you have
[math] \sum_{n=0}^\infty (-1)^n (1-\varepsilon)^n = \dfrac{1}{2-\varepsilon} = \dfrac{1}{2} + O(\varepsilon) [/math]
and more generally
[math] \sum_{n=0}^\infty (-1)^n (s-\varepsilon)^n = \dfrac{1}{1+s} + O(\varepsilon) [/math]
You can visualize the limit if you plot [math] \dfrac{1}{1-z} [/math] over C. (that's a dare, I can't find the plot anymore on the web)
Somewhat similarly, the sequence
[math] a_n := \sin \left ( n + \frac{1}{2} \right) [/math] is
+1, -1, +1, -1, +1, ...
and so while
[math] \sum_{n=0}^\infty \sin \left ( n + \frac{1}{2} \right) [/math]
doesn't converge, you may regularize it and arive at
[math] \sum_{n=0}^\infty \sin \left ( (1 + \frac {2i} {\pi} \varepsilon) (n + \frac{1}{2}) \right) = \dfrac {1} {2} e^\varepsilon [/math]
>>7944402
Not necessarily. The two humps do cancel each other out if you start at zero and go to infinity but you could just as easily start -pi/2 or pi/2 or any value for that matter and have a non-zero answer. There's and infinite amount of answers.
>>7944402
[math] \lim_{x\to \infty } \, \sin (x)=[-1,1] [/math]
