[Math] A Ramanujan-like summation: is it correct? Is it extensible

divergent-seriesramanujan-summationsequences-and-seriessummation

I'm still exercising with summation-procedures which I try to make correct Ramanujan-summations. I'm looking at the (gap-)series
$$ s(1/2,2) = (1/2)^1+(1/2)^{4}+(1/2)^{9}+(1/2)^{16}+(1/2)^{25}+… $$
and more general at
$$ s(b,p) = b+b^{2^p}+b^{3^p}+b^{4^p}+b^{5^p}+… \tag 1$$
For convergent case where $0<b<1$ we can evaluate this approximately using serial evaluation of expression (1), say
$$ \begin{eqnarray} s(1/2,1) &=& 1 \\s(1/2,2) & =& 0.564468413… \\
s(1/2,3)&=&0.503906257… \end{eqnarray}$$.

On the other hand, expanding that formal sum in a double series of a series of formal exponential series in $x$ and collecting like powers of the argument $x$ I got something like Ramanujan-summation (where I only express the Bernoulli-numbers in the Ramanujan formula by the equivalent zeta-references at negative arguments).
So I define the expressions
$$ \begin{eqnarray} I(b,p)&=& \int_{t=- 1}^\infty b^{(1+t)^p} dt \\
Z(b,p)&=& \sum_{k=0}^\infty \beta^k {\zeta(-pk) \over k!} \qquad \text{ where } \beta = \log(b) \end{eqnarray} \tag 2$$
Then the summation-method $\mathcal Q$
$$ s(b,p) \underset{\mathcal Q}= I(b,p) + Z(b,p) \tag 3$$
gives for the above convergent cases $s(1/2,1),s(1/2,2)$ the correct results by numerical approximations. (The second case is convergent because each zeta at negative even argument is zero).
$$ \begin{eqnarray}
I(1/2,1) &=& + 1.44269504089… \\
Z(1/2,1) &=& -0.44269504089…\\
s(1/2,1) &\underset{\mathcal Q}=& \phantom + 1\\
\hline\\
I(1/2,2) & =& + 1.06446701943… \\
Z(1/2,2)&=& -0.5\\
s(1/2,2)&\underset{\mathcal Q}=& \phantom + 0.56446701943… \end{eqnarray}$$.

Also various examples with other parameters, where everything in the $\mathcal Q$-method is convergent seems to approximate the results taken by serial evaluation of (1) correctly.
For the case $s(1/2,3)$ the sum $Z(1/2,3)$ in (2) is no more convergent. However, using a Noerlund-summation $\mathcal N$ for that expression I could approximate the expected result (taken by evaluation of (1)) satisfyingly.
$$ \begin{eqnarray}
I(1/2,3) & =& + 1.00901976692… \\
Z(1/2,3)&\underset{\mathcal N}=& -0.50561…\\
s(1/2,3)&\underset{\mathcal Q}=& \phantom + 0.50340…\\
\hline \\
s(1/2,3)&=& \phantom + 0.503906257 \text{ by serial summation } \end{eqnarray}$$.

So perhaps this allows more generalization.

Q1: Is this eq (2) a valid reconstruction of the Ramanujan-summation, at least in principle?

Q2: The integration-bounds were experimentally. Are they correct? and if: how could I have derived them correctly?

[update1]: There must be some systematical error. For all even $p=2q$ the sum of the zetas $Z(1/2,2 \cdot q) =-1/2 = \zeta(0) $ and the correct sum $s(1/2,2\cdot q) = 1/2 + \delta $ is always bigger than $1/2$. So it is required that $$I(1/2,2\cdot q)= s(1/2,2\cdot q) – Z(1/2,2 \cdot q) = 1 +\delta$$
But now, by Wolframalpha I see that $$ \int_{t=-1}^\infty 1/2^{(1+t)^p} dt = { \Gamma(1+1/p)\over \sqrt[p]{\log 2} } $$ which decreases below $1$ at $p \approx 3.44395$ so for all $p=4,6,8,…$ the $\mathcal Q$-summation cannot hold.
Hmm…

[update 2,update 3] Here are some data which show very nice approximation for the $\mathcal Q$-summation with the serial summation for small p of $p=0.1 \ldots 1.2$
with errors $ \lt 1e-100$.
$$
\small \begin{array} {r|rrl}
p & s(1/2,p) & err &=s(1/2,p) – (I(1/2,p)+Z(1/2,p) \\
\hline
0.1 & 141745219.752 & -4.00E-156 \\
0.2 & 749.678969635 & 2.82E-167 \\
0.3 & 31.0876588123 & 4.18E-166 \\
0.4 & 7.95318949339 & 1.33E-164 \\
0.5 & 3.78821923065 & -1.96E-162 \\
0.6 & 2.37977624581 & 9.16E-160 \\
0.7 & 1.72998968339 & -2.28E-156 \\
0.8 & 1.37118947539 & -6.79E-152 \\
0.9 & 1.14893798357 & -4.14E-146 \\
1.0 & 1.00000000000 & 5.69E-139 \\
1.1 & 0.89438296309 & -1.18E-128 \\
1.2 & 0.81625996055 & 2.25E-114 \\
&&& \text{very small errors – perhaps is $\mathcal Q$-summation valid? }\\
\vdots \\
1.7 & 0.61712884341 & 2.28 E-18 & \text{using Borel-summation}\\
1.8 & 0.59639864943 & -1.27 E-11 \\
1.9 & 0.57905393685 & -0.0000000385 \\
&&& \text{errors increase- $\mathcal Q$-summation tends to become invalid? }\\
\vdots\\
3.0 & 0.503906257 \phantom {00} & 0.00040176 & \text{Z(1/2,p) by Noerlund/Borel-summation }
\end{array}
$$
Because of the very slow convergence at the small p I used the Pari/GP-function sumpos which seems to be able to approximate the true value avoiding the computation of millions of terms in the serial summation (1). For the parameter $p \gt 1.5$ I crosschecked the Noerlund-summation for $Z(1/2,p)$ by Borel-summation.

Best Answer

A non rigorous approach, just simple "intuitive methode". Because the power is $x^2$, the period is 2. Make sure to include the zero point in it's polynomal notation. I will later try to write it for all power's of $x$.

$$ \sum_{x=1}^{\infty} (c)^{-x^2}$$ $$ \sum_{x=1}^{\infty} \sum_{n \in \mathbb{Z}}\ (\ln(c)x^2)^n(-1)^n/n!=\sum_{x=1}^{\infty} \sum_{n \in \mathbb{Z}}\ \frac{(x)^n (\ln(c))^{n/2}}{(n/2)!} (e^{i\pi n/2})\sum_{k=0}^1 \frac{e^{\frac{2i \pi*kn}{2}}}{2}$$

$$= \sum_{n \in \mathbb{Z}}\ \frac{\zeta(-n) (\ln(c))^{n/2}}{(n/2)!} \frac{(e^{i\pi n/2}+e^{3i\pi n/2})}{2}$$

First take the limit as n goes to -1. Which gives us the value $\frac{ \sqrt{\pi}}{2 \sqrt{\ln(c)}}$

There's another value at the "normal" order, mainly as $n$ goes to 0. At $n=0$ the value is $-\frac{1}{2}$.

Notice that both answers give a very very good approaximation for the sum, but are a very little off, (orders like $10^-{6}$, really small). Mathematically I can't fully show why (yet). More on intuition based, let $h$ be a very small value, and you have rewrite the function you are trying to sum as $g(n)f(n)$ with $g(n)$ being a periodically zero, with an order $h$ near the zero so $g(n+h)$ and $f(n)$ being a function that grows more/equal then a constant value, the regularised (as in the upper or lower limit is infinity) the sum is not that well defined and you just have to be careful. Lots of horrible talk and explainations I soon will be ashamed of so I am hoping for a better explaination. So here's what I found to be the solution.

If $n$ is even, the zero's don't have any influence for the sum and can be regarded as 0.

If n is uneven this isn't the case: $$= \sum_{n \in \mathbb{Z}}\ \frac{\zeta(-n) (\ln(c))^{n/2}}{(n/2)!} e^{i\pi n/2} \frac{(1+(-1)^n)}{2}$$

use that $$\zeta(-n)=\frac{\zeta(n+1) n! (i^{n+1}+(-i)^{n+1})}{(2\pi)^{n+1}} $$ $$\frac{n!}{(n/2)!2^n}=\frac{((n-1)/2)!}{\sqrt{\pi}}$$

$$ \sum_{n \in \mathbb{Z}}\ \frac{\zeta(n+1)(\ln(c))^{n/2}\big(\frac{(n-1)}{2}\big)! }{\pi^{n+3/2}} \frac{(e^{i\pi n/2}+e^{3i\pi n/2}))((e^{i\pi (n+1)/2}+e^{3i\pi (n+1)/2}))}{4}=$$

We only had to look at when $n=2m-1$.

$$ \sum_{m \in \mathbb{Z}}\ \frac{\zeta(2m)(m-1)! (ln(c))^{m-1/2} }{2\pi^{2m+1/2}} i(e^{3i\pi m}-e^{i\pi (m)})(-1)^{m}$$ Because it's summed over all integers including the negative ones, we subsitute m=-m.

$$ \sum_{m \in \mathbb{Z}} \frac{\zeta(-2m)(-m-1)! \pi^{2m-1/2}}{2(\ln(c))^{m+1/2}} i(e^{-3i\pi m}-e^{-i\pi (m)}) (-1)^{m}$$

I've had the alternating/peridiocally zero's kinda horrible defined. But approach the limit of m=m+h at the zero's/infinities to look at the (uneven) zero's. So that $e^{-3i\pi (m+h)}-e^{-i\pi (m+h)}=-2hi\pi$ I know you should work this better out with it's period, and notice that everything is defined in the same "complex" period, maybe use trigonometry functions, but those functions are less intunitive. Again it's non rigorous, only to show that there's some way to a good solution with this methode. And know that $(-n-1)!=h^{-1}/n!$ if n is an integer $$ \sum_{m \in \mathbb{Z}} \frac{\zeta(-2m)(-m-1)! \pi^{2m+1/2}}{(\ln(c))^{m+1/2}} (h) (-1)^{m}$$ $$ \frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{m \in \mathbb{Z}} \frac{\zeta(-2m) \pi^{2m}}{m!(\ln(c))^{m}} (-1)^{m}$$ Officially I'd say you'd had to write it again out also as (and maybe you couldn't ignore the other zero's, I just don't know yet, it just works, I've to polish it obviously)

$$ \frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{m \in \mathbb{Z}} \frac{\zeta(-m) \pi^{m}}{(m/2)!(\ln(c))^{m/2}} (-1)^{m/2}(1+(-1)^m)$$ $$ \frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}\sum_{m \in \mathbb{Z}} \frac{x^{m} \pi^{m}}{(m/2)!(\ln(c))^{m/2}} (-1)^{m/2}(1+(-1)^m)$$ Which is the regularised sum of $$\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{ln(c)}}$$

$$\sum_{x=1}^{\infty} (c)^{-x^2}=\frac{\sqrt{\pi}}{2 \sqrt{ln(c)}}-1/2+\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{\frac{-\pi^2 x^2}{\ln(c)}}.$$

The first two value's being your I(c^(-1),2) and Z(c^(-1),2) and the sum being your systematically error. And i can only agree that the systematically error is something counter intunitive at first, but also arise when you apply this summation methode to other series as i recall well known easier examples such as $\sum_{x=1}^{\infty} \frac{1}{2x^2-1}$.

In similair rough fashion for $$ \sum_{x=1}^{\infty} (c)^{-x^d}$$

$$ \sum_{m \in \mathbb{Z}}\ \frac{(-1)^{1/d}\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \sum_{k=0}^{d-1} \frac{e^{\frac{2ipi*km}{d}}}{d}$$

$$\sum_{m \in \mathbb{Z}}\ \frac{\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{e^{i\pi m/d}(e^{2 i \pi m}-1)}{d(e^{ 2i \pi m/d}-1)}$$

Now the value at m=0, will always be -1/2. at limit m goes to -1 gives:

$$\frac{-\pi}{\sin(\pi/d) (\ln(c))^{1/d}(-\frac{1+d}{d})!}=\frac{(1/d)!}{\ln(c)^{1/d}}$$

Which is similair to the integeral you stated, if d is not even you got to indeed sum is over the other value's and don't forget the error.

A try of mine to find the error:

$$\sum_{m \in \mathbb{Z}} \frac{\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{e^{i\pi m/d}(e^{2 i \pi m}-1)}{d(e^{ 2i \pi m/d}-1)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{-\zeta(-m)(ln(c))^{m/d}}{(m/d)!} \frac{h \pi}{d \sin(m \pi /d)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{-\zeta(m+1) m!(ln(c))^{m/d}}{(2 \pi)^{m+1} (m/d)!} \frac{h \pi (i^{m+1}+(-i)^{m+1})}{d \sin(m \pi /d)}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(m+1) m!(ln(c))^{m/d}}{(2 \pi)^{m} (m/d)!} \frac{h \pi (sin(m \pi/2)}{d \sin(m \pi /d)}$$

Use that:

$$\frac{m!}{(m/d)!}= d^{m+1/2} (2 \pi)^{(1-d)/2} \prod_{j=1}^{d-1} \big(\frac{m-j}{d}\big)!$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(m+1) (ln(c))^{m/d}}{} \frac{h (sin(m \pi/2)}{\sin(m \pi /d)}\frac{d^{m-1/2} (2 )^{(3-d)/2-m} (\pi)^{(1-d)/2-m}\prod_{j=1}^{d-1} \big(\frac{m-j}{d}\big)!}{}$$

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1) (ln(c))^{-m/d}}{} \frac{h (sin(-m \pi/2)}{\sin(-m \pi /d)}\frac{d^{-m-1/2} (2 )^{(3-d)/2+m} (\pi)^{(1-d)/2+m}\prod_{j=1}^{d-1} \big(\frac{-m-j}{d}\big)!}{}$$

Split up for the zero's in $\prod_{j=1}^{d-1} $

$$\sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1) (ln(c))^{-m/d}}{} \frac{h (sin(-m \pi/2)}{\sin(-m \pi /d)}\frac{d^{-m-1/2} (2 )^{(3-d)/2+m} (\pi)^{(3-d)/2+m}}{ \prod_{j=1}^{d-1}\sin(\frac{\pi (m+j)}{d}) \big(\frac{m+j}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{} \frac{h (sin((-dm-k) \pi/2)}{\sin((-dm-k) \pi /d)}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(3-d)/2+dm+k}}{ \prod_{j=1}^{d-1}\sin(\frac{\pi (dm+j+k)}{d}) \big(\frac{ \pi (dm+j+k)}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{(sin((-dm-k) \pi/2)}{\sin((-dm-k) \pi /d)}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d}) \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=0)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d})}\frac{d^{(-dm-k)-1/2} (2 )^{(3-d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

$$\prod_{(j=0)\wedge j\neq d-k}^{d-1}\sin(\frac{\pi (dm+j+k)}{d})=d2^{1-d}(-1)^{?}$$ I don't have the minus sign for now. Will come back at that later, assuming it's not there cause i've already enough in one equation.

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{}\frac{d^{(-dm-k)-3/2} (2 )^{(1+d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{ \prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}$$

to make it readable $$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!}\frac{(2 )^{(1+d)/2+dm+k} (\pi)^{(1-d)/2+dm+k}}{d^{(dm+k)+3/2} }$$

$$\sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k) (ln(c))^{(-dm-k)/d}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} \frac{(\frac{2\pi}{d})^{dm+k+1/2} (\frac{2}{\pi})^{d/2}}{d}$$

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(dm+k)}}{m!} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)\wedge j\neq d-k}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} $$

I am already suprised the formula above seem to work for d=2. If only the m didn't sneaked into the product life would be good. Kinda used this is a note pad, i will clean it up one day, but results count.

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}} \sum_{k=1}^{d-1}\frac{\zeta(-dm+1-k)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(dm+k)}}{} \frac{\sin((dm+k) \pi/2)}{\prod_{(j=1)}^{d-1} \big(\frac{dm+j+k}{d}-1\big)!} $$

$$\frac{\sqrt{\frac{2\pi}{d}} (\frac{2}{\pi})^{d/2}}{d} \sum_{m \in \mathbb{Z}/d\mathbb{Z}} \frac{\zeta(-m+1)}{} \frac{ \bigg(\frac{2\pi}{d\sqrt[d]{ln(c)}}\bigg)^{(m)}}{} \frac{\sin((m) \pi/2)}{\prod_{(j=1)}^{d-1} \big(\frac{m+j}{d}-1\big)!} $$

It shows how I think you can proceed (I let it rest, as it got messy, and i probably made some errors, and made things unnessecary hard), with a doable solution of d=2, as was the original question. Maybe you can or should add $m \in d\mathbb{Z}$, to make it smooth, Again i do not know yet.

Related Solutions

[Math] Interpretation of Ramanujan summation of infinite divergent series

There are answers in -as I think- in mathoverflow for "what are physical interpretations of zeta at negative arguments" (or the like, I don't have the actual words in mind). Just browse a bit through MO or MSE using that terms as filter.

0) Need of a new concept - and a basic requirement: Make the general term explicite

For the interpretation and assignment of a meaningful value in your series you must define, how the general term, say $a_k$, shall be constructed, i.e. how the value of $a_k$ is dependend on its index $k$. This is a requirement which shall be demanded in all discussions of divergent series (and should not been forgot when such series, written only in their "obvious" numerical forms, are considered at all).

In this example the value of the general term, say $a_k$, reduces simply to the index $k$ if taken from $k=1 \ldots \infty $ and we have actually the formal statement $S = \sum_{k=1}^\infty k$. But because the summing procedure is obviously divergent we must introduce some new concept if we want assign something meaningful to it.

1) Geometric series

A useful idea here was (for instance already by L. Euler) that we can cofactor a variable $x$ at the terms $a_k$ such that for some $x$ the series becomes convergent and evaluatable to some finite value and see what happens with the sequence of resulting sum-values, when the expressions of $x$ approximate $1$ . For instance, we could redefine your sum as $$ S(x) = 1 + 2x + 3x^2 + 4x^3 + ... \\ S = \lim_{x \to 1^-} S(x) $$ because there are some $x$ (actually it is important, that there exists a continuous interval of $x$) where this is convergent, for instance for $x=0 \ldots \frac12$. We can then even observe that with that definition the sum $S(x)$ has a closed form $$S(x) = {d\over dx}{1\over 1-x } = {1\over (1-x)^2 }$$ This is already a very nice representation, because it allows now even negative $x$ which result in finite values for the alternating sum, even if divergent, because it is accepted to assign that fraction's value also to its formal power series-expression even if the latter is divergent - except, well, except if $x=1$.
But what we want here is actually just the latter, so we have not yet a satisfying answer.

2) Dirichlet series

Another idea is to apply a function of $x$ to the exponent of the terms $a_k$ and then let $x$ go to $1$ . Thus we consider the notation $$ S(x) = 1^x + 2^x + 3^x +4^x + ... $$ which is convergent for a continuous interval of $x$, for instance of $x = -\infty \ldots -2$ (and even a bit more). But still, for $x=1$ we get no obvious answer.
But interestingly, for the whole interval of convergence we have also the functional relation $$ \begin{matrix} \text{let } &A(x) &=& 1^x - 2^x + 3^x - 4^x + \ldots \\ \text{then } &S(x) &=& A(x) + 2 \cdot 2^x \cdot S(x) \\ \text{and} & S(x) &=& A(x)/(1-2 \cdot 2^x) \end{matrix} $$ And now we can assign a meaningful value to $S = \lim_{x \to 1^- }$. Either by evaluating $A(1)$ as conditionally converging series in the given form, or by the above geometric-series interpretation and its derivative at $x=-1$ where both ways of evaluation give the same rational expression $A(1)=\frac 14$.
After that nothing more divergent is there and we get $$ S = \lim_{x \to 1^-} S(x) = \frac 14 / (1-2\cdot 2^x) = \frac 14 / (1-4)= - \frac 1{12} $$ having now a proposal for a meaningful assignment of a finite value for the infinite divergent series.

3) Caveat

Of course that value must conform with all and any cases where such series occur in mathematics, and one example of a contradiction with known results taken from the mathematic without divergent summation would invalidate that procedure!
Moreover, we would hope that even in the physical world, where we model some observations with that series, such an assignement of values to a divergent series would hold. Interestingly such observations exist in the real world and it seems that the whole process and also the final value meets the modeling of that observations. (Examples are given to the according questions either in MSE here or in mathoverflow (I've just found (1)(2)), you can do a search for the important words "zeta" and "at negative arguments" or similar)

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.

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.

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