Grandi's series with 3 numbers So it appears that given any number X, where S = X-X+X-X..., the sum (S) is always X/2.
However, what about if we have a series like the following:
3-2-1 + 3-2-1 + 3-2-1 ...
How would one go about calculating this?
 A: Here's one summation method: let $(a_n)_{n\geq0}$ be a bounded sequence of real (or complex) numbers. We define
$$S\bigl((a_n)_{n\geq0}\bigr)=\lim_{x\to1^-}\sum_{n=0}^{+\infty}a_n x^n$$
(if this limit exists). Note that since we assumed that the sequence $(a_n)_{n\geq0}$ is bounded, the radius of convergence of the power series $\sum_n a_n z^n$ is at least $1$, so that the limit makes sense (but may not exist).
In the case of Grandi's series: it is well-known that for $x\in(-1,1)$,
$$\sum_{n=0}^{+\infty}(-1)^n x^n=\frac1{1+x}\underset{x\to1^-}\to\frac12,$$
so that $S\bigl(((-1)^n)_{n\geq0}\bigr)=1/2$.
Now define the sequence $(a_n)_{n\geq0}$ as
$$\forall p\geq0,\ \begin{cases}a_{3p}=3\\a_{3p+1}=-2\\a_{3p+2}=-1.\end{cases}$$
The sequence $(a_n)_{n\geq0}$ is obviously bounded, so the corresponding power series has a radius of convergence non-less than one. For $x\in(-1,1)$,
$$\sum_{n=0}^{+\infty}a_n x^n=3\sum_{p=0}^{+\infty}x^{3p}-2\sum_{p=0}^{+\infty}x^{3p+1}-\sum_{p=0}^{+\infty}x^{3p+2}=\frac{3-2x-x^2}{1-x^3}=\frac{x+3}{x^2+x+1}\underset{x\to1^-}\to\frac43.$$
So $4/3$ seems to be a legitimate value for your sum. Note that with Cesaro's summation we would obtain the same answer.
A: Notice the partial sums are periodic:
$$\begin{align}S_1&=3\\S_2&=1\\S_3&=0\\\vdots\ &\phantom{mn}\vdots\end{align}$$
Thus, the Cesaro sum is the average of the first period of terms:
$$S=\frac{3+1+0}3=\frac43$$
Which is in agreeance with gniourf_gniourf's answer.
