Summation method for $1-3+9-27 + \dots $ Is there a summation method to compute the series
$$1-3+9-27+\dots ? $$
We knew this is divergent in the usual sense, but there are summation methods for giving a sense to divergent series. For example: Cesaro summation, Abel summation, Euler summation, and so on. I am looking for a method like that.
I am thinking about the Euler transformation of this series, which gives the series
$$\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\dots$$
But this series does not converge in the usual sense. Hence, it is not Euler summable.
Thank you!
 A: *

*Any linear stable summation method under which $\sum_{n=0}^{\infty} (-3)^n$ is summable must yield
$$ \sum_{n=0}^{\infty} (-3)^n = \frac{1}{4}. $$
This is because the sum can be computed only by appealing to both linearity and stability: if $S = \sum_{n=0}^{\infty} (-3)^n$ is summable, then
\begin{align*}
S
&= 1 + \sum_{n=0}^{\infty} (-3)^{n+1} \tag{stability} \\
&= 1 + (-3)S \tag{linearity}
\end{align*}
and thus solving this equation for $S$ yields $S = \frac{1}{4}$. Unfortunately, it is not Abel summable.

*On the other hand, it is Borel summable:
$$
\mathcal{B}\sum_{n=0}^{\infty} (-3)^n
= \int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{(-3)^n}{n!}t^n \right) e^{-t} \, dt
= \int_{0}^{\infty} e^{-4t} \, dt
= \frac{1}{4}. $$

*The answer may change if we replace the underlying field from $\mathbb{C}$ to other complete field $F$ of characteristic zero (which makes sense because $F$ contains an isomorphic copy of $\mathbb{Q}$). For instance, if we pick $F = \mathbb{Q}_3$ as the $3$-adic field, then $\sum_{n=0}^{\infty} (-3)^n$ converges absolutely to $\frac{1}{4}$ under the $|\cdot|_3$-norm.
A: It's a geometric series. If you want to assign a finite, real number to characterise your series, then I believe that
$$
\frac{1}{1-(-3)}=\frac14
$$
is the most natural value to assign.
A: This is divergent. Each term switches the sign and roughly triples the absolute value. This is not summable
A: By right it’s a divergent series but anyway, assign the series a value $S$ such that
$$S=1-3+9-27+\ldots$$
So,
$$S=1-3S$$
Hence,
$$S=\frac14$$
