Cesáro sums and the actual limit

1.3k Views Asked by At

My textbook, as an aside, defines the Cesáro sum as follows: $$ \sigma_n= \frac{s_1+...+s_n}{n}= \frac{1}{n}\sum_{k=1}^ns_k, $$ where $$ s_n = \sum_{k=1}^na_k. $$ This method is used, I am told, to find a value of otherwise divergent sums. For exemple the sum $1-1+1-1+1-1+1-1+...$ becomes $1/2$. The notation $(C,2)$ means that you have calculated the Cesáro sum of a Cesáro sum. Let's say you have a sum that doesn't have a finite value until you calculate $(C,10)$, is $(C,10)$ the actual limit?

Is the Cesáro sum the limit of the series or simply a value that coincides with the limit for convergent series?

2

There are 2 best solutions below

0
On BEST ANSWER

The limit of a convergent series coincides with its [first] Cesáro sum (proof) and consequently, with its higher order Cesáro sums as well.

In particular, the contrapositive states that if $(C,1)$ does not exist, then the series does not converge.

0
On

The answer you accepted is accurate, but more can be said. In short $(C,10)$ is not the limit but it could be called something like the (tenth order) Cesaro limit. And the series is not summable (in the usual sense) and does not have a (usual) limit. But it is Cesaro summable (of order $10$ if you wish.)

First let's move from series to sequences. From a series $ \sum a_i$ We have a a related series $s_1,S_2,\cdots$ of partial sums. If this sequence converges to a value $v$ according to the usual definition we say that the series is a summable series or convergent series. That value is the sum or limit and we might say $v=(C,0). $ Otherwise the series is divergent.

So by definition a divergent series does not have a limit. There are good reasons for those conventions/definitions. But we can also define the Cesaro transform which converts $s_1,s_2,\cdots$ series to $\sigma_1,\sigma_2,\cdots$ and if this new sequence has a limit we say that the original series is Cesaro-summable with Cesaro sum $(C,1).$ (I'm sticking to your notation, although I am not fond of it.)

Note that $(C,2)$ would not be the Cesaro sum of the Cesaro sum but rather the limit of the Cesaro transform of the Cesaro transform of the sequence $s_1,s_2,\cdots.$ We could apply the Cesaro transform to sequences which arose in other ways.

If a series is summable it is also Cesaro summable and $(C,1)=(C,0)$. Sometimes when the convergence to $(C,0)=v$ is slow, the convergence of the Cesaro transform to $C(,1)=v$ is much more rapid.

The whole subject is undeniable cool. How useful it is, I don't know.