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There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$ and computing the expansion of $n'(\xi)$ at $\xi = 0$, we obtain $$\frac {n'(\xi)} x = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} \sim \int_{-\infty}^\infty n'(\xi) e^{x/e - x \xi^2} d\xi = e^{x/e} \left( \sqrt{\frac{2 \pi x} e} - \frac 1 {12} \sqrt{\frac {\pi e} {2 x}} + O(x^{-3/2}) \right), \quad x \to \infty.$$

There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$, computing the expansion of $n'(\xi)$ at $\xi = 0$ and extending the integration range to $(-\infty, \infty)$, we obtain $$\frac {n'(\xi)} x = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} = \int_{-\infty}^\infty x \left( \sqrt{\frac 2 e} - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + O(\xi^4) \right) e^{x/e - x \xi^2} d\xi = \\ \sqrt{\frac \pi 2} \,e^{x/e} \left( 2\sqrt{\frac x e} - \frac 1 {12} \sqrt{\frac e x} + O(x^{-3/2}) \right), \quad x \to \infty,$$ which gives $\ln f(x)$ with an error of order $O(x^{-2})$.

There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$ and computing the expansion of $n'(\xi)$ at $\xi = 0$, we obtain $$\frac {n'(\xi)} x = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} \sim \int_{-\infty}^\infty n'(\xi) e^{x/e - x \xi^2} d\xi \sim e^{x/e} \left( \sqrt{\frac{2 \pi x} e} - \frac 1 {12} \sqrt{\frac {\pi e} {2 x}} + O(x^{-3/2}) \right), \quad x \to \infty.$$

There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$ and computing the expansion of $n'(\xi)$ at $\xi = 0$, we obtain $$\frac {n'(\xi)} x = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} \sim \int_{-\infty}^\infty n'(\xi) e^{x/e - x \xi^2} d\xi = e^{x/e} \left( \sqrt{\frac{2 \pi x} e} - \frac 1 {12} \sqrt{\frac {\pi e} {2 x}} + O(x^{-3/2}) \right), \quad x \to \infty.$$

There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$ and computing the derivative of the inverse function at $\xi = 0$, we obtain $$g(\xi) = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} \sim x \int_{-\infty}^\infty g(\xi) e^{x/e - x \xi^2} d\xi \sim e^{x/e} \left( \sqrt{\frac{2 \pi x} e} - \frac 1 {12} \sqrt{\frac {\pi e} {2 x}} + O(x^{-3/2}) \right), \quad x \to \infty.$$

There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$ and computing the expansion of $n'(\xi)$ at $\xi = 0$, we obtain $$\frac {n'(\xi)} x = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} \sim \int_{-\infty}^\infty n'(\xi) e^{x/e - x \xi^2} d\xi \sim e^{x/e} \left( \sqrt{\frac{2 \pi x} e} - \frac 1 {12} \sqrt{\frac {\pi e} {2 x}} + O(x^{-3/2}) \right), \quad x \to \infty.$$