How to prove that Grandi's series $= \frac{1}{2}$ using Euler transform Let $x$ denote Grandi's series $1-1+1-1+1-1+1-...$ This implies that $$ x = 1\text{ or}\\\ x = 0\text{ or}\\\ 1-x = 1 - (1-1+1-1+1-...) = x \implies 2x = 1 \implies x = \frac{1}{2}$$ Where the last sum seems counterintuitive as $\sum_{k=0}^n (-1)^k = \frac{1}{2} (-1)^{n}+1 \neq \frac{1}{2}$ but is proven by taking the Cesàro or Abel sum. Is it possible to calculate in favour of $x=\frac{1}{2}$ by using Euler transform?

You might want to be more cautious about " _x_ "; it's not clear what that letter is supposed to mean here. Anyway, yes, the Euler transform of the series $1-1+1-1+\cdots$ is $\frac12+0+0+\cdots$, which converges to $\frac12$. So the _Euler sum_ of $1-1+1-1+\cdots$ is $\frac12$.

As Euler himself put it:

> I. _Sit igitur proposita haec series Leibnitzii:_ $$S=1-1+1-1+1-1+\mathrm{\&c.}$$ in qua cum omnes termini fint aequales, fient omnes differentiae $=0$, ideoque ob $a=1$, erit $S=\frac12$.