MHB Sum of infinite divergent series

[Amer]

It is well known that the below series are divergent

$1 - 1 + 1 - 1 + \cdots $

$1 - 2 + 3 - 4 + \cdots $

$1 + 2 + 3 + \cdots $

But after i watched a video in youtube for the channel " Numberphile " they proved that the first is equal to 1/2 , 1/4 and the last one is -1/12 !

The way to find such thing. Let

$S = 1 - 1 + 1 + \cdots $ Then

$S = 1 - (1 -1 + 1 - \cdots ) = 1 - S \rightarrow 2S = 1 \rightarrow S = \frac{1}{2}$

The second
$ S_2 = 1 - 2 + 3 -4 + \cdots $
$ S_2 + S_2 = 1 + ((-2 + 1) + ( 3 - 2) + \cdots ) = 1 + ( -1 + 1 -1 +\cdots ) = 1 - S $ But $S = \frac{1}{2}$
$2 S_2 = \frac{1}{2} $ Hence $S_2 = \frac{1}{4}$.

The question is what is the matter a divergent series has a sum ? is all divergent series has a value in some way ?

 

[chisigma]

It is well known that the below series are divergent

$1 - 1 + 1 - 1 + \cdots $

$1 - 2 + 3 - 4 + \cdots $

$1 + 2 + 3 + \cdots $

But after i watched a video in youtube for the channel " Numberphile " they proved that the first is equal to 1/2 , 1/4 and the last one is -1/12 !

The way to find such thing. Let

$S = 1 - 1 + 1 + \cdots $ Then

$S = 1 - (1 -1 + 1 - \cdots ) = 1 - S \rightarrow 2S = 1 \rightarrow S = \frac{1}{2}$

The second
$ S_2 = 1 - 2 + 3 -4 + \cdots $
$ S_2 + S_2 = 1 + ((-2 + 1) + ( 3 - 2) + \cdots ) = 1 + ( -1 + 1 -1 +\cdots ) = 1 - S $ But $S = \frac{1}{2}$
$2 S_2 = \frac{1}{2} $ Hence $S_2 = \frac{1}{4}$.

The question is what is the matter a divergent series has a sum ? is all divergent series has a value in some way ?

If the series 1 - 1 + 1 - 1 +... diverges, then writing S= 1 - 1 + 1 - 1 +... is a nonsense because no real number S exists that satisfies such a requirement...

Kind regards

$\chi$ $\sigma$

 

[Amer]

If the series 1 - 1 + 1 - 1 +... diverges, then writing S= 1 - 1 + 1 - 1 +... is a nonsense because no real number S exists that satisfies such a requirement...

Kind regards

$\chi$ $\sigma$

So it is converge?. but can't we say it is undefined ( similar to 0/0 or $\infty$ )since if we

S = 1+ ( -1 + 1 ) + ( -1 + 1 ) + ... = 0

S = (1 - 1 ) + ( 1 - 1 ) + ... = 1

and by the previous way we get 1/2. So it is has three values maybe more. And each value came from logical true steps. Is there any problem ?

 

[alyafey22]

This is quoted from mathworld

"The Riemann series theorem states that, by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge"

Look at his link.

This is due to the existence of other definitions of convergence other than Cauchy convergence. The first series is called the Cesàro summation. The series is clearly not convergent according to Cauchy criteria but the series has a value which is $\frac{1}{2}$. That value contradicts our common sense of addition and subtraction because a summation of integers should result in an integer. Indeed the set of integers is not dense in the real line. There are actually other criterion of summations like Abel summation and Lindelöf summation and there definitions are different than the usual definition of convergence.

Here are more links

https://en.wikipedia.org/wiki/Cesàro_summation
Abel summability | planetmath.org

Both prove that the first series is Cesàro and Abel summable.

 

[chisigma]

So it is converge?. but can't we say it is undefined ( similar to 0/0 or $\infty$ )since if we

S = 1+ ( -1 + 1 ) + ( -1 + 1 ) + ... = 0

S = (1 - 1 ) + ( 1 - 1 ) + ... = 1

and by the previous way we get 1/2. So it is has three values maybe more. And each value came from logical true steps. Is there any problem ?

On this occasion I am afraid that the first 'logical step' is not true ...

... more precisely if You have a series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ and You demonstrate that it converges, i.e. is $\displaystyle \lim_{k \rightarrow \infty} \sum_{n=1}^{k} a_{n}$ exists and it is equal to S, then You can write $\displaystyle \sum_{n=1}^{\infty} a_{n} = S$...

... but if You the original series diverges, You are not authorized to do that and the first 'logical step' is false...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

It is well known that the below series are divergent

$1 - 1 + 1 - 1 + \cdots $

$1 - 2 + 3 - 4 + \cdots $

$1 + 2 + 3 + \cdots $

But after i watched a video in youtube for the channel " Numberphile " they proved that the first is equal to 1/2 , 1/4 and the last one is -1/12 !

Those series are all divergent, of course. But interpreting them in ways that appear to lead to values for their sum can sometimes lead to genuine mathematical or physical insights. See this Wikipedia page, which explains how Ramanujan derived and used his formula "$1+2+3+4+\ldots = -\frac1{12}$". Ramanujan knew that his use of this formula could easily be misinterpreted. In a letter to Hardy, he wrote "If I tell you this you will at once point out to me the lunatic asylum as my goal." But a recent author described this equation as "one of the most remarkable formulae in science".

Playing with divergent series in this way is dangerous, and you should not attempt it at home. But in the hands of an expert it can be an important heuristic tool.

 

[Amer]

Playing with divergent series in this way is dangerous, and you should not attempt it at home. But in the hands of an expert it can be an important heuristic tool.

This remind me of WWE show hehe. Do not try this at home.