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For what (if any) $d\in\mathbb{N}$ does the following expression have a finite value?

$$u(d)=\sum\limits_{m=-\infty}^{\infty}\sum\limits_{n=0}^{\infty}\frac{1}{2^{d(2n+|m|)}}\binom{2n+|m|}{n}^d$$

When I attack it for $d=2$, I can see that the $m=0$ case will be $\sum\limits_{n=0}^\infty \frac{(2n)!^2}{(n!)^4 16^n}$ which diverges if formula $(8)$ in this question is correct (Wolfram Alpha agrees). Thus I think $u(d)$ diverges for $d=2$ but I'm not sure how to proceed for larger values of $d$ (for $d=3$ the $m=0$ case is evaluated as $\frac{\pi}{\Gamma(\frac{3}{4})^4}$ but Wolfram Alpha doesn't help me with the whole double sum); I assume that asymptotic approximations will have to be used.

For the context of the question, it is very similar to a question I recently asked here. In the present case, I was attacking the following problem (just for fun): 'If $d$ people stand next to each other in a line and play a game where at each step each of them tosses a fair coin and for each person if he throws heads then he moves forwards $1$ metre and if tails he moves backwards $1$ metre; what is the probability $p(d)$ that they will ever be in a line next to each other again?'.

Using a similar technique (i.e. using generating functions, from the method used in this paper), I got $p(d)=1-\frac{1}{u(d)}$ with $u(d)$ as above. Thus for $d$ such that $u(d)$ diverges $p(d)=1$, and if $u(d)$ converges, then if I am correct it would have to be in $[1,\infty)$ in order for $p(d)$ to be a probability. I would like to know whether this is the case, and for what $d$ will $u(d)$ diverge?

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  • $\begingroup$ It would make sense to study how these points spread as the time goes on, then conclude the wanted probability is zero or one by Kolmogorov's theorem and Cantelli's inequality. $\endgroup$ – Jack D'Aurizio Jan 13 '17 at 8:55
  • $\begingroup$ @JackD'Aurizio But the wanted probability need not only be zero or one? Polya's constants take on all sorts of values. $\endgroup$ – Anon Jan 14 '17 at 2:00

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