What is the Ramanujan summation for the series $\sqrt[n]{2}$ A Ramanujan summation is a

technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series

In my case, I'm interested in assigning a value to the divergent series
$$\sum_{n=1}^\infty f(n) \ \ \ \ \ \ \ \text{where}\ \ \ \ f(n)=\sqrt[n]{2}$$
According to the Wikipedia page (and my understanding), the Ramanujan summation is
$$\sum_{n=1}^\mathfrak{R} f(n)=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N f(n)-\int_{1}^N f(t)dt\Bigg]$$
Thus
$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg]$$
Taking the antiderivative
$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(\ln2\Big(\text{li}\ 2-\text{Ei}\frac{\ln2}{N}\Big)+N\sqrt[N]{2}-2\Bigg)\Bigg]$$
Moving some constants outside the limit
$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=2-\ln2\cdot\text{li}\ 2+\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(N\sqrt[N]{2}-\ln2\cdot\text{Ei}\frac{\ln2}{N}\Bigg)\Bigg]$$
It's at this point I'm unsure of how to proceed. I'm not terribly confident what the limit converges to. From my computational estimates up to $N=10^8$, I find that
$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}\approx1.6$$
But due to floating point errors or slow convergence, it deviates substantially enough for me to not be confident about any more digits.
I'd like to know if this converges at all, and if it does, is there a (reasonably) closed form / relation to other constants?
 A: I'm not sure I'm calculating the same thing, and as regularizations of divergent sums need not be unique, I will ask the OP to do some more numerical work on his formula.  The following is how I would find a closed-form expression, 'in the manner of Ramanujan summation.'
$$F_n(x) := \sum_{k=1}^n x^{1/k} = \frac{n}{n}\sum_{k=1}^n (x^{n/k})^{1/n} \to n \int_{0}^1 x^{1/(nt)} dt$$
as a Riemann sum as $n \to \infty.$  This is the first term on the right-hand side of your defining equation
$$ \sum_{k=1}^{\mathcal{R}} f(k) = \lim_{n \to \infty} \Big(\sum_{k=1}^n f(k) - \int_{1}^n f(t) dt \Big)$$
and has an explicit closed-form
$$ F_n(x) = n \ x^{1/n} + \Gamma(0,\frac{\log(x)}{n} )\log(x) . $$  I've used the incomplete gamma function, but the first argument tells us it will reduce to the Ei integral.  By the way, this is a perfectly good asymptotic approximation, giving 4 significant figures for $x=2$ and $n=1000.$  The second term also has an explicit expression,
$$ \int_1^n x^{1/t} dt = -x + n\ x^{1/n} + \log{x} \Big(-\text{Ei}(\log{x}/n)+ \text{li}(x) \Big).$$
(Both integrals have been performed in Mathematica.)  Now subtract and do an asymptotic expansion.  The singular parts cancel and you are left with
$$ 
\sum_{k=1}^{\mathcal{R}} x^{1/k} = -\big(x+ \log{(x)}\, \text{li}(x) \big) $$
where li$(x)$ = LogIntegral[x] in Mathematica. When I plug in $x=2,$ I get the value of -2.72445.  This is very different from the OP's answer, with not even sign agreement.  However, we should not forget the Ramanujan summation is closely aligned with zeta regularization, and with it you can 'prove' an infinite sequence of positive numbers is a negative number, e.g.,
$$ 1+2+3+... = \lim_{s \to \  -1} \sum_{k=1}^\infty k^{-s}  = \lim_{s \to \  -1}\zeta(s) = -\frac{1}{12} $$
I'm wary of regularization, though it does lead to some cute results.  Physicists are sometimes forced into such shenanigans with perturbation theory.
