$a = 1 + 1 + 1 + \cdots$
$a + a + a \cdots =$
$(1 + 1 + 1 + 1 \cdots) +$
$\quad \space \space ~ (1 + 1 + 1 \cdots) +$
$\quad \quad \space \space ~ \space \space ~ (1 + 1 \cdots) +$
$\quad \quad \quad \space \space ~ \space \space ~ \space \space ~ (1 + \cdots) + \cdots$
Thus: $a+ a+ a + \cdots = 1 + 2 + 3 + 4\cdots$.
Since $a + a + a + \cdots = a (1 + 1 + 1 \cdots) = a^2$, we thus have that $1 + 1 + 1 \cdots = \pm \frac{i}{\sqrt{12}}$.