I wanted to verify that
$$ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +\cdots = 1 $$
Using Ramanujan summation.
Recall that the Ramanujan summation formula is that
$$ f(1) + f(2) + f(3) + \cdots = - \frac{f(0)}{2} + i \int_0^{\infty} \frac{f(is)-f(-is)}{e^{2\pi s} - 1} ds $$
Here was have $f(n) = 2^{-n}$
So the result should show that:
$$ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = -\frac{1}{2}+ i \int_{0}^{\infty} \frac{2^{-is}-2^{is}}{e^{2\pi s} - 1} ds$$
I evaluated the RHS in Wolfram Alpha and surprisingly! I do not get an answer of $1$.
See: here
Instead I get a result that is very close to $-0.44$ with some change.
Why might this be so? Is it because of internal floating point error on Wolfram's end, or does Ramanujan summation just not sum some convergent series? (And if Ramanujan summation doesn't sum some series: what is the significance of their implied R-sum values?)