Limit of an expression that includes the Lambert W function

Question

What is $\lim\limits_{k\rightarrow\infty}\ \sqrt{-(k+1)W_{-1}\left(\frac{-1}{(1+\frac{1}{k})\exp\left(\frac{1+\left(\frac{2}{k}\right)\ln\left(\frac{\sqrt{k}\Gamma(k/2)}{\sqrt{2}\Gamma((k+1)/2)}\right)}{\left(1+\frac{1}{k}\right)}\right)}\right)-k}$ ?

Here, $\Gamma(\cdot)$ is the usual gamma function; and $W_{-1}(\cdot)$ is a real branch of the Lambert $W$ function: https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf .

I suspect, based on numerical calculations from Python code as shown below, that the answer might be close to $1.55 \approx\sqrt{1+\sqrt{2}}$.

`from scipy.special import lambertw
 from scipy.special import gamma
 import math
 import numpy as np
 import matplotlib.pyplot as plt


 n=200
 f=np.empty(n)
 for k in np.arange(1,n+1,1):
   r1=(1+(1/k))
   r2=1+(2/k)*math.log(math.sqrt(0.5*k)*gamma(k/2)/gamma((k+1)/2))
   r3=r2/r1
   z=-1/(r1*np.exp(r3))
   w=-lambertw(z,-1)
   f[k-1]=math.sqrt((k+1)*w-k)

 fig, ax = plt.subplots()
 ax.set(xlabel='k', ylabel='f(k)')
 ax.grid()
 ax.plot(np.arange(1,n+1,1), f,'r')`

Answer 1

This seems to be an excellent exercise of composition of Taylor series.

We have $$\sqrt{-(k+1) W_{-1}\left(-\frac{1}{A}\right)-k}$$

Using Stirling approximation and continuing with Taylor series, we have $$A=e+\frac{e}{k^2}-\frac{7 e}{6 k^3}+O\left(\frac{1}{k^4}\right)$$ $$-\frac{1}{A}=-\frac{1}{e}+\frac{1}{e k^2}-\frac{7}{6 e k^3}+O\left(\frac{1}{k^4}\right)$$ Now, close to $-\frac 1e$ we have $$W_{-1}(x) \sim -1-\sqrt{2(1+ex)}\implies W_{-1}\left(-\frac{1}{A}\right)=-1-\frac{\sqrt{2}}{k}+\frac{7}{6 \sqrt{2} k^2}+O\left(\frac{1}{k^3}\right)$$ $$-(k+1) W_{-1}\left(-\frac{1}{A}\right)-k=\left(1+\sqrt{2}\right)+\frac{5}{6 \sqrt{2} k}+O\left(\frac{1}{k^2}\right)$$ $$\color{red}{\sqrt{-(k+1) W_{-1}\left(-\frac{1}{A}\right)-k})=\sqrt{1+\sqrt{2}}+\frac{5}{12 \sqrt{2 \left(1+\sqrt{2}\right)} k}+O\left(\frac{1}{k^2}\right)}$$

Edit

Having confessed in a comment my passion for composition of Taylor expansions, I decided to go further in order to obtain much more than the limit itself.

So, we have $$-\frac 1 A=-\frac{1}{e}+\frac{1}{e k^2}-\frac{7}{6 e k^3}+\frac{2}{3 e k^4}-\frac{1}{20 e k^5}-\frac{119}{360 e k^6}+\frac{391}{630 e k^7}-\frac{359}{315 e k^8}+\frac{55639}{45360 e k^9}+\frac{18833}{21600 e k^{10}}+O\left(\frac{1}{k^{11}}\right)$$ For the expansion of Lambert function, I used the expansion $$W_{-1}(z)=\sum_{i=0}^9 b_i \,p^i \qquad \text{where} \qquad p=-\sqrt{2(1+e z)}$$ (the $b_i$'s being given here - just below equation $(37)$).

So, now we have $$-(k+1) W_{-1}\left(-\frac{1}{A}\right)-k=(1+\sqrt 2)+\frac{8+5 \sqrt{2}}{12 k}-\frac{96+99 \sqrt{2}}{864 k^2}-\frac{256-73 \sqrt{2}}{17280 k^3}+\frac{30983 \sqrt{2}-11264}{829440 k^4}+\frac{4820992+2119625 \sqrt{2}}{69672960 k^5}+O\left(\frac{1}{k^{6}}\right)$$ Comparing lhs and rhs shows that the relative error is smaller than $0.002$% a soon as $k >4$.

Notice that pushing the expansion drastically changes the coefficient of $\frac 1k$.

Update

I do not remember in which paper were given the bounds

$$-1-\sqrt{2u}-u < W_{-1}(-e^{-u-1}) < -1-\sqrt{2u}-\frac{2}{3}u$$ $$-\log (A)-\sqrt{2} \sqrt{\log (A)-1}<W_{-1}\left(-\frac{1}{A}\right)< -\frac{2 \log (A)}{3}-\sqrt{2} \sqrt{\log (A)-1}-\frac{1}{3}$$ Now, using $$A=e+\frac{e}{k^2}-\frac{7 e}{6 k^3}+O\left(\frac{1}{k^4}\right)$$ $$(1+\sqrt2)+\frac{8+5 \sqrt{2}}{12 k}<-(k+1) W_{-1}\left(-\frac{1}{A}\right)-k<(1+\sqrt2)+\frac{12+5 \sqrt{2}}{12 k}$$ and then the limit using the squeeze theorem.