I am having difficulty understanding how to find the Cesaro sum of the series:
$1-1+0+1-1+0+\dots$
I know the sequence of partial sums will be:
$1,0,0,1,0,0,1,0,0,1,0,0,\dots$
And hence the average of the partial sums will be:
$1, 1/2, 1/3, 1/2, 2/5, 1/3, 3/7,3/8, 1/3, 2/5, 4/11, 1/3, \dots$
I know this is basically all I have to do, but I'm confused as to actually see what the sum converges to.
How can I show what the this converges to?
It looks like it is just $1/3$, but how can I be sure and show that?
Another question:
I have this series and another one that were both generated by inserting zeroes into the Grandi series.
This obviously changes the Cesaro sum of this series from the Grandi series (we can see above this isn't converging on $1/2$ as the Grandi series does).
My question is, given a convergent series (a standard convergent series), will its sum change if we insert zeroes into it?
My thinking so far is that it shouldn't.
When summing a normal series, you don't need to take averages so the addition of zeroes won't change anything.
You just keep adding zeroes to the partial sum, but they won't have any effect, since you don't have to take an average as you have to in the Cesaro sum.
Does this seem correct at all?
Thank you
In the $k$th group of three terms you have the values $$\frac{k}{3k-2},\ \frac{k}{3k-1},\ \frac{k}{3k}$$ where the "group index" $k$ runs through $1,2,3,\cdots .$ In this form it is easier to see that as $k$ gets large the limit is $1/3$ as you say.
For the other question, if a series is originally convergent, insertion of $0$ terms doesn't change things, for essentially the reason you gave. But if the original series is not convergent in the usual sense, yet cesaro-converges, then as your example shows insertion of zeros can change the cesaro sum.