I am working through the last bit of a random walk proof to show that a 3-d random walk is transient. The result I am looking for states that:
$\frac {1}{2}^{2s} {{2s}\choose{s}} \sum_{j+k\leq{n}} (\frac {1}{3^s} \frac{(s)!}{k!j!(s-j-k)!})^2$ $= \frac {c}{n^{3/2}}$
I can understand that the above can be manipulated to get $\frac {1}{2}^{2s} {{2s}\choose{s}} (\frac {1}{3^s} \frac{s!}{(n/3)!^3})$ and thence $\frac {1}{2}^{2s} {{2s}\choose{s}} (\frac {1}{3^s} \frac{\sqrt{2n\pi}e^{-n}n^n}{(\sqrt{2s\pi} e^{(-s/3)}(\frac{s}{3})^{s/3})^3})$ by Stirling's approximation.
By distributing the powers and simplifying, I obtain $(1/2)^{2s} {{2s}\choose{s}} \frac{\sqrt{2\pi s}}{(\sqrt2\pi s/3)^3}$ which I can then get into the form $\frac{2\frac {1}{2}^{2s}}{2\sqrt{\pi/3}^3s^{2/3}}$ by applying Stirling again to the ${2s}\choose{s}$ term.
However, I would have though that the continued presence of n in the $1/2^{(2s)}$ would have meant that we can't refer to the numerator as a constant c?