# Talk:Cesàro summation

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## Sentence doesn't make sense

The following sentence in the article doesn't make sense: "In fact, any series which diverges to (positive or negative) infinity the Cesàro method also leads to a sequence that diverges likewise, and hence such a series is not Cesàro summable."

Should it say something along the lines of "In fact, any series whose partial sums diverge to plus or minus infinity also leads to a sequence that diverges likewise in the Cesàro sense."?

## Pun

"...Cesàro summation is an alternative means of assigning a sum...." Ha ha, "means". I hope the pun was intentional. 128.255.45.80 (talk) 23:43, 21 March 2010 (UTC)

## difference between Cesàro sum and limit?

The article currently says that the generalized Cesàro sum can be restated as

${\displaystyle (C,\alpha )-\sum _{j=0}^{\infty }a_{j}=\lim _{n\to \infty }\sum _{j=0}^{n}{\frac {n \choose j}{n+\alpha \choose j}}a_{j}.}$

However, the article also says that if the series converges then the Cesàro sum is equal to the limit, and that the generalized Cesàro sum is just an iterated version of the normal Cesàro sum – this would seem to imply that if ${\displaystyle \sum _{j=0}^{\infty }a_{j}}$ exists then the difference on the left-hand side should be zero. Am I missing something? Joriki (talk) 21:15, 26 November 2010 (UTC)

The LHS is not a difference. It's (bad but standard) notation for the sum. Sławomir Biały (talk) 12:24, 4 February 2011 (UTC)
Currently, the ASCII hyphen is being automatically converted to a minus sign because it's inside [itex] tags. Perhaps this would be less confusing if we were to use a non-math hyphen or an en-dash or something? Joule36e5 (talk) 23:54, 1 February 2013 (UTC)
I changed it to an en dash. Joule36e5 (talk) 09:20, 26 February 2015 (UTC)

## Irregular oscillator

"Since a sequence that is ultimately monotonic either converges or diverges to infinity, it follows that a series which is not convergent but Cesàro summable oscillates." Note that it doesn't have to be a regular cycle... For example:
n = {1,2,3,4,5,6,7,8,...} integers greater than 0
x_n = cos(n)+cos(n sqrt(2)) cycle that never repeats because of impossible factoring

This is a trivial example. Since by definition, a transcendental number never repeats, this means that it is impossible to have harmonics with n and n sqrt(2). Since the natural period of the cosine function is a multiple of Pi, some care has to be taken when constructing such examples to not allow the terms inside the cosine functions have certain values (or patterns). n and n*1.41... with n being positive integers satisfies this requirement. 71.196.246.113 (talk) 09:56, 19 January 2012 (UTC)

Sorry, meant to mention my question. Does anyone think this is important enough of a point to add to the article. And I dare someone to prove that the square root of 2 is rational!  ;) (It's an old disproof in any decent book on proofs)71.196.246.113 (talk) 10:01, 19 January 2012 (UTC)

## Example of the Grandi series

I do read quite often exactly what I read here about the cesàro mean of the Grandi series, yet I never found a plausible explanation. This article states the following:

${\displaystyle \lim _{n\to \infty }t_{n}=1/2.}$

with ${\displaystyle t_{n}={\frac {1}{n}}\sum _{k=1}^{n}s_{k}}$

From my understanding (and the German Wiki Article) however, this is false. Starting from the limit: ${\displaystyle \lim _{n\to \infty }t_{n}}$ I think it should become:

${\displaystyle \lim _{n\to \infty }t_{n}=\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}s_{k}.=0}$

Dreikommavierzehn (talk) 22:14, 12 May 2015 (UTC)

The German article de:Cesàro-Mittel (permanent link) is incorrect. For more details on the Cesàro sum of Grandi's series, including a citation, see Summation of Grandi's series#Cesàro sum. Melchoir (talk) 23:12, 12 May 2015 (UTC)
The explanation from the link makes sense, yet I don't see the correctness in the above example. The limit of 1/n * something will still be 0. At the very least, I think that the statement written with a limit is flawed. 2001:628:2010:22:14DB:2B64:CC76:577A (talk) 08:42, 13 May 2015 (UTC)
Yes, if the series ${\displaystyle \sum _{k=1}^{\infty }s_{k}=\lim _{n\to \infty }\sum _{k=1}^{n}s_{k}}$ converged, then the limit ${\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}s_{k}}$ would be 0. That's the sense in which the limit of 1/n * something is 0. However, in this case, the "something" by itself, ${\displaystyle \sum _{k=1}^{\infty }s_{k}}$ does not converge. It diverges to positive infinity, and ${\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}s_{k}=1/2}$ measures how quickly it does so. Melchoir (talk) 20:06, 13 May 2015 (UTC)
Alright. But now it looks like ${\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}s_{k}=}$ "0∗∞" an indeterminate form. Aside from the arguments provided by Summation of Grandi's series#Cesàro sum, I'm under the impression that the argumentation always boils down to some form of reordering the partial sums. Which clearly cannot be done. Dreikommavierzehn (talk) 22:22, 14 May 2015 (UTC)
That's not how indeterminate forms work. I don't wish to sound rude, but if you're looking for a fuller explanation, you might get a better response at Wikipedia:Reference desk/Mathematics. Melchoir (talk) 03:52, 15 May 2015 (UTC)
Thank you, I'll hit my books. Dreikommavierzehn (talk) 11:02, 17 May 2015 (UTC)