Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing $k$ so long as $2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for $k$ (and making approximations valid for large $N$) gives \begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\ &< \frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+x^2}}{4}N\2N-\sqrt{x^4+4x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+4x^2}}{4}N\\ \end{align} and so for large $N$ the largest term will be $k=f(x)N$, with $f(x)=\frac{2+x^2-\sqrt{x^4+x^2}}{4}$. f(x)=\frac{2+x^2-\sqrt{x^4+4x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any$\epsilon$we have arbitrarily large$N$such that this is within$\epsilon$of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y) we get \begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\ \end{align} and so (note we make the approximation\log(N-1)\approx \log(N)) \begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &\approx -\frac{\log(2)}{2}-f(x)-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ \end{align} thus \begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\ \end{align} which should be quite close to the actual value due to the accuracy of Stirling's for large values. 4 deleted 2 characters in body LetT_N(x)$denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem$p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression$k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$ and so we continue to make the denominator smaller by increasing$k$so long as$2x^2(k+1)(N-1)<(N-2k-1)(N-2k)$. Solving for$k$(and making approximations valid for large$N) gives \begin{align}k &< \frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\ &< \frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{1+x^2-\sqrt{x^4+x^2}}{4}N\2N-\sqrt{x^4+x^2}N}{4}=\frac{2+x^2-\sqrt{x^4+x^2}}{4}N\\ \end{align} and so for largeN$the largest term will be$k=f(x)N$, with$f(x)=\frac{1+x^2-\sqrt{x^4+x^2}}{4}$. f(x)=\frac{2+x^2-\sqrt{x^4+x^2}}{4}$. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any $\epsilon$ we have arbitrarily large $N$ such that this is within $\epsilon$ of an integer. Thus $$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$ and so taking logarithms and applying Stirling's approximation ($\log(y!)\approx y\log(y)-y$) we get \begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\ \end{align} and so (note we make the approximation $\log(N-1)\approx \log(N)$) \begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &\approx -\frac{\log(2)}{2}+f(x)-2-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\ \frac{\log(2)}{2}-f(x)-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ \end{align} thus \begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\ \end{align} which should be quite close to the actual value due to the accuracy of Stirling's for large values.
Let $T_N(x)$ denote the largest term of the sum. Note that $$\frac{\log T_N(x)}{N}\leq \frac{\log (Z_N(x))}{N}\leq \frac{\log ((N/2)T_N(x))}{N}=\frac{\log T_N(x)}{N}+\frac{\log(N/2)}{N}$$ and that the limits of both sides are identical, so by Squeeze theorem $p(x)=\lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}$. To find the largest term, we4 want to minimize the expression $k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}$. Note that $$\frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k-2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{(k+1)(N-1)}{2x^2(N-2k-1)(N-2k)}$$ \frac{(k+1)!(N-2k-2)!(N-1)^{k+1}(\sqrt{2}x)^{2k+2-N}}{k!(N-2k)!(N-1)^k(\sqrt{2}x)^{2k-N}}=\frac{2x^2(k+1)(N-1)}{(N-2k-1)(N-2k)}$$and so we continue to make the denominator smaller by increasing k so long as (k+1)(N-1)< 2x^2(N-2k-1)(N-2k). 2x^2(k+1)(N-1)<(N-2k-1)(N-2k). Solving for k (and making approximations valid for large N) gives$$\begin{align}k &< \frac{-1+N-4 x^2+8 N x^2-\sqrt{1-2 N+N^2-24 x^2+8 N x^2+16 N^2 x^2+16 x^4}}{16 x^2}\frac{N x^2+2N-x^2-1-\sqrt{N^2x^4+4N^2x^2-2Nx^4+2Nx^2+x^4-6x^2+1}}{4}\\ &< \frac{N+8 N x^2-\sqrt{N^2+16 N^2 x^2}}{16 x^2}=\frac{1+8x^2-\sqrt{1+16x^2}}{16 x^2}N\frac{N x^2+ 2N-\sqrt{x^4+x^2}N}{4}=\frac{1+x^2-\sqrt{x^4+x^2}}{4}N\\ \end{align}$$and so for large N the largest term will be k=f(x)N, with f(x)=\frac{1+8x^2-\sqrt{1+16x^2}}{16 x^2}. f(x)=\frac{1+x^2-\sqrt{x^4+x^2}}{4}. Note that we can assume equality (rather than accuracy to the nearest integer) without hurting our calculations since the limit exists and for any \epsilon we have arbitrarily large N such that this is within \epsilon of an integer. Thus$$T_N(x) = 2^{-N/2}\frac{N!}{(f(x)N)!((1-2f(x))N!)}(N-1)^{-f(x)N}(\sqrt{2}x)^{(1-2f(x))N}$$and so taking logarithms and applying Stirling's approximation (\log(y!)\approx y\log(y)-y) we get$$\begin{align}\log(T_N(x)) &= -\frac{N}{2}\log(2)+\log(N!)-\log((f(x)N)!)-\log((1-2f(x))N!)\\ & \;\;\;\;-f(x)N\log(N-1)+(1-2f(x))N\log(\sqrt{2}x)\\ &\approx -\frac{N}{2}\log(2)+N\log(N)-N-f(x)N\log(f(x)N)+f(x)N\\ & \;\;\;\;-(1-2f(x))N\log((1-2f(x))N)+(1-2f(x))N-f(x)N\log(N-1)\\ & \;\;\;\;+(1-2f(x))N\log(\sqrt{2}x)\\ \end{align}$$and so (note we make the approximation \log(N-1)\approx \log(N))$$\begin{align} \frac{\log(T_N(x))}{N}&\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x)N) \\ & \;\;\;\;-(1-2f(x))\log((1-2f(x))N)-f(x)\log(N-1)\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &\approx -\frac{\log(2)}{2}-f(x)+\log(N)-f(x)\log(f(x))-f(x)\log(N) \\ & \;\;\;\;-(1-2f(x))\log(1-2f(x))-(1-2f(x))\log(N)-f(x)\log(N-1)\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ &\approx -\frac{\log(2)}{2}+f(x)-2-f(x)\log(f(x))-(1-2f(x))\log(1-2f(x))\\ & \;\;\;\;+(1-2f(x))\log(\sqrt{2}x)\\ \end{align}$$thus$$\begin{align} p(x) &= \lim\limits_{N\to\infty} \frac{\log(T_N(x))}{N}\\ &\approx -\frac{\log(2)}{2}-f(x)(1+\log(f(x)))+(1-2f(x))(\log(\sqrt{2}x)-\log(1-2f(x)))\\ \end{align}\$ which should be quite close to the actual value due to the accuracy of Stirling's for large values.