Grandi's series with 3 numbers

Question

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?

Answer 1

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.

Answer 2

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.