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Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably $$ \rho = \frac{729(2+3\alpha)^2}{64(1+\alpha)^2} \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha} \approx 47.7322896460174547. $$

The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.

To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.

ADDED: The stuff inside the large parens simplifies to 1, and the stuff outside them simplifies to $$ \rho = \frac{6561}{16} - \frac{452709}{64}\alpha - \frac{1358127}{256}\alpha^2,$$ which implies that $\rho$ is the smallest zero of $$ 1048576\rho^3 - 4353564672\rho^2 + 4518872583696\rho - 205891132094649. $$

Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably $$ \rho = \frac{729(2+3\alpha)^2}{64(1+\alpha)^2} \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha} \approx 47.7322896460174547. $$

The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.

To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.

ADDED: The stuff inside the large parens simplifies to 1, and the stuff outside them simplifies to $$ \rho = \frac{6561}{16} - \frac{452709}{64}\alpha - \frac{1358127}{256}\alpha^2,$$ which implies that $\rho$ is the smallest zero of $$ 1048576\rho^3 - 4353564672\rho^2 + 4518872583696\rho - 205891132094649,$$ which also happens to be $3^{10}/2^8$ times the smallest zero of $y^3-18y^2+81y-16$, and I guess that's about as much simplification as possible.

Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably $$ \frac{729(2+3\alpha)^2}{64(1+\alpha)^2} \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha} \approx 47.7322896460174547. $$ Can anyone simplify that?

The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.

To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.

ADDED: The stuff inside the large parens simplifies to 1, and the stuff outside it simplifies to $$ \frac{6561}{16} - \frac{452709}{64}\alpha - \frac{1358127}{256}\alpha^2.$$

Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably $$ \rho = \frac{729(2+3\alpha)^2}{64(1+\alpha)^2} \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha} \approx 47.7322896460174547. $$

The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.

To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.

ADDED: The stuff inside the large parens simplifies to 1, and the stuff outside them simplifies to $$ \rho = \frac{6561}{16} - \frac{452709}{64}\alpha - \frac{1358127}{256}\alpha^2,$$ which implies that $\rho$ is the smallest zero of $$ 1048576\rho^3 - 4353564672\rho^2 + 4518872583696\rho - 205891132094649. $$

Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably $$ \frac{729(2+3\alpha)^2}{64(1+\alpha)^2} \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha} \approx 47.7322896460174547. $$ Can anyone simplify that?

The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.

To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.

Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably $$ \frac{729(2+3\alpha)^2}{64(1+\alpha)^2} \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha} \approx 47.7322896460174547. $$ Can anyone simplify that?

The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.

To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.

ADDED: The stuff inside the large parens simplifies to 1, and the stuff outside it simplifies to $$ \frac{6561}{16} - \frac{452709}{64}\alpha - \frac{1358127}{256}\alpha^2.$$