Convergence of $\sqrt{\frac{2N}{d}}\frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}$? (expected value from random walks)

Question

I saw from this post (stack) that the expected distance from the origin after $N$ steps in $d-$dimensional space is $$ \sqrt{\frac{2N}{d}}\frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}. $$ I was curious to know the convergence behaviour of this (as $d\rightarrow\infty$ or as $N\rightarrow\infty$), I was hoping somebody here could help me find out?

The context for this is that I have the function $$ F_N(x) = \sum_{p\leq N}\log pe^{2\pi ipx} $$ (where $p$ is a prime number), and the average value of the absolute value of this function squared is $N\log N + o(N\log N)$; i.e. $$ \int_{0}^{1}|F_N(x)|^2dx = N\log N + o(N\log N). $$ In a sense summing $N$ random numbers from the unit circle can be bounded above by $N\log N$. I'm wondering whether the expected value from a random walk could give some insight into whether this bound is very crude or not. In fact it seems likely that it is, because creating a better bound for it would allow the circle method to dig deeper into Goldbach's conjecture.

Answer 1

If $N\to\infty$, the expression diverges. The other limit is more interesting. If I recall correctly, it is the case that $$\Gamma(d+a) \sim \Gamma(d)\,d^a\qquad\operatorname{as}d\to\infty.$$ Hence, $$\sqrt{\frac{2N}{d}}\frac{\Gamma(d/2+1/2)}{\Gamma(d/2)}\sim \sqrt{\frac{2N}{d}}\sqrt\frac{d}{2}=\sqrt{N}\qquad\operatorname{as}d\to\infty.$$ Thus, we see that $$\lim_{d\to\infty}\sqrt{\frac{2N}{d}}\frac{\Gamma(d/2+1/2)}{\Gamma(d/2)} = \sqrt{N}.$$

Answer 2

The only dependence on $N$ is the factor of $\sqrt{N}$, so there's no convergence issues there. In $d$ you can use Sterling's formula to deduce that

$$\sqrt{\frac{2N}{d}} \frac{\Gamma\left(\frac{d+1}{2}\right)}{\Gamma\left(\frac{d}{2}\right)} \sim \sqrt{N} \times \left(1 - \frac{1}{4d} + \frac{1}{32 d^2} + \frac{5 }{128d^3} - \frac{21 }{2048d^4} - \frac{399}{8192 d^5} + \frac{869}{65536 d^6} + \ldots\right)$$