Is $\lim_{n\to\infty}2^{-n/2}\sum_{k=1}^{n-1}\frac{k^2\cdot2^{k/2}}{k+\frac{1}{2}}\left(\frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})}\right)^2=1+\sqrt{2}$?

Question

After I've read a question posted on MathOverflow I tried to combine it with the statement of a problem published in the American Mathematical Monthly (that I've omitted, in fact in my sum are dropped two terms since the Gamma function is undefined at such points). That is, the MONTLHY published a theorem that now I've specialized for some sequeces and numbers, and after I've calculated with Wolfram Alpha online calculator my conjecture, the following in next question.

Question. Prove or refute $$\lim_{n\to\infty}2^{\frac{-n}{2}}\sum_{k=1}^{n-1}\frac{k^2\cdot2^{\frac{k}{2}}}{k+\frac{1}{2}}\left(\frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})}\right)^2=1+\sqrt{2},\tag{1}$$ where $\Gamma(s)$ denotes the gamma function. Many thanks.

I don't know if from the detailed and interesting solution of MONTHLY's problem can be deduced my closed-form $(1)$ easily (I do not think so). Thus my conjecture is based on experiments with the mentioned online tool and my code

sum 2^(-100000000/2) k^2 2^(k/2)/(k+1/2) (Gamma(k)/Gamma(k+1/2))^2, from k=1 to 100000000-1

or clik over More digits after you run this code

sum 2^(-1000000000/2) k^2 2^(k/2)/(k+1/2) (Gamma(k)/Gamma(k+1/2))^2, from k=1 to 1000000000-1

I know that should be useful Stirling's approximation or some inequality related to the gamma function, combined with some numeric method of summation.

References:

[1] The sum of an hydrogen atom related infinite series, posted on MathOverflow with quote to the article by Tamar Friedmann and C. R. Hagen, Quantum Mechanical Derivation of the Wallis Formula for $\pi$, Journal of Mathematical Physics 56, 112101 (2015).

[2] Problem E 1760 [1965,183] A Convergent Sequence Arising from a Difference Equation, proposed by I. I. Kolodner, American Mathematical Monthly Vol. 73, No. 4, (1966), pages 414-415.

Answer 1

By Stolz-Cesaro Theorem, $$\lim_{n\to\infty}2^{-n/2}\sum_{k=1}^{n-1}\frac{k^2\cdot2^{k/2}}{k+\frac{1}{2}}\left(\frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})}\right)^2= \lim_{n\to\infty}\frac{1}{2^{(n+1)/2}-2^{n/2}}\frac{n^2\cdot2^{n/2}}{n+\frac{1}{2}}\left(\frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\right)^2\\ =(\sqrt{2}+1)\lim_{n\to\infty}n\left(\frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\right)^2.$$ Now use the this approximation property of the Gamma function: for $\alpha\in\mathbb{C}$, $$\lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{n^{\alpha}\Gamma(n)}=1.$$