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

I've not yet seen that definition of Ramanujan-summation as in your first formula.
But for testing I've tried your third formula $$ \sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg] \tag 3 $$ implemented in Pari/GP for high precision.

First, using W|A I've got the following expression for the integral, setting $ß=\log(2)$ $$ \int \exp( \frac ßt) dt = t\exp(\fracßt) - ß \text{Ei}(\fracßt) + const \tag {4.1} $$ and $$ \begin{align}I(N)&= \int_{t=1}^{N} \exp( \frac ßt) dt \\ &= (N\exp(\fracßN) - ß \text{Ei}(\fracßN))&-(1\exp(\fracß1) - ß \text{Ei}(\fracß1)) \\I(N) &=N\exp(\fracßN) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß)) \end{align} \tag {4.2}$$ With this I come for some not too large N in the near of your found value: $$K(N) = \sum_{k=1}^N \exp( \fracßk) \tag{4.2}$$ and $$S(N) = K(N) - I(N) \tag {4.3}$$

  N       S(N)        
   100  1.60585777814
  1000  1.60273245539
 10000  1.60242047744
100000, 1.60238928520

It seems to converge, and to the value that you gave in your OP. But doing the serial summation in $K(N)$ to even higher N (to get more accuracy) is time-consuming, so I reconstruct this series such that Pari/GP can calculate this easier when $N$ is in the billions... This is the rewriting as a double series, where the expression for each $\sqrt[k] 2$ is expanded in the exponential-series on $\fracßk$

$$ \begin{array} {} \text{lhs} & =\text{rhs1} &+ \text{rhs2} \\ \hline \exp(ß/1) & = 1+ß/1 & +ß^2/1^2/2! &+ß^3/1^3/3! & + \cdots \\ \exp(ß/2) & = 1+ß/2 & +ß^2/2^2/2! &+ß^3/2^3/3! & + \cdots \\ \exp(ß/3) & = 1+ß/3 & +ß^2/3^2/2! &+ß^3/3^3/3! & + \cdots \\ \vdots & \vdots \\ \exp(ß/N) & = 1+ß/N & +ß^2/N^2/2! &+ß^3/N^3/3! & + \cdots \\ \vdots & \vdots \\ \end{array} \tag 5$$ Looking at the column-sums we see, that the first two columns $\text{rhs1}$ give divergent series, but the following columns $\text{rhs2}$ are convergent. So we operate in two parts: evaluate the RHS2 for the limit $N \to \infty$ immediately by the sum of zetas $$ \text{RHS2}(\infty) = ß^2/2! \zeta(2) + ß^3/3! \zeta(3) + ß^4/4! \zeta(4) + ... \approx 0.473841903568 \tag {5.1} $$ Since the columnsums in RHS1 diverge we reformulate the columnsums up to the N'th partial sum in immediate values in terms of $N$ and of harmonic-numbers $H(N)=\psi(1+N)+\gamma$ (or H(N)=psi(1+N)+Euler in Pari/GP): $ \text{RHS1}(N)=N+ ß \cdot H(N)$ so $$ \begin{array} {}K(N) &= N + ß \cdot H(N) &+ 0.473841903568\\ I(N) &= N\exp(\fracßN) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß)) \\ S(N) &= K(N)&- I(N) \end{array} \tag {6}$$

giving convergence to the value

  N    S(N)           difference S(N_{k+1})-S(N_k)
 1000: 1.60297258955 
10^5 : 1.60239168746 -0.000580902092269
10^6 : 1.60238640626 -0.00000528119790414
10^12: 1.60238581946 -0.000000586799480429
10^24: 1.60238581946 -5.86800097238 E-13
10^48: 1.60238581946 -5.86800097239 E-25

So I think, your own approximation went into the right way.

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P.s. In eq.6 the divergence can be reduced by cancellation; the exponential in $I(N)$ be expanded to $N(1+ß/N) + O(1/N)=N + ß + O(1/N)$ and the divergence in $N$ cancels with that $N$ in $K(N)$. Next, the expression $\text{harm}(N)$ can be rewritten as $\log(N)+\gamma$ for $N \to \infty$ and be put together with the $\text{Ei()}$ - expression in $I(N)$ getting $$ \lim_{N\to \infty} \text{Ei}(ß/N) + \log(N) = \gamma + \log(\log(2)) + O(1/N)$$ leading to the shortened version of eq.6 $$ \begin{array} {} \lim_{N \to \infty} S(N) &= & ß \cdot (\log(N)+\gamma) &+ 0.473841903568\\ &&- ( ß + O(1/N) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß))) \\ &=& ß\gamma -ß+2 -ß\text{Ei}(ß) &+ 0.473841903568\\ &&+ ß (\text{Ei}(\fracßN)+\log(N) ) \\ &=& 2+ß(2\gamma + \log(ß) -1 -\text{Ei}(ß)) &+ 0.473841903568\\ &=& 1.60238581946 \end{array} \tag {6a}$$ where in the second-last line the divergence $N$ has been cancelled and a constant expression emerged.