Return to Answer

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} \mathrm{e}^{ - nt} \mathrm{d}t} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {\mathrm{e}^{ - xp(t)} q(t) \mathrm{d}t} , $$ with $p(t) = - t\mathrm{e}^{ - t}$ and $q(t)=\mathrm{e}^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim \mathrm{e}^{z} \sqrt {2\pi z} \sum\limits_{n = 0}^\infty {\frac{{a_n }}{{z^n }}} = \mathrm{e}^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/\mathrm{e}$ and $$ a_n = \frac{1}{{2^n n!}}\left[ {\frac{{\mathrm{d}^{2n} }}{{\mathrm{d}t^{2n} }}\left( {\mathrm{e}^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)\mathrm{e}^{ - t} }}} \right)^{n + 1/2} } \right)} \right]_{t = 0} . $$ An alternative expression for the coefficients involving the Bernoulli polynomials $B_k(x)$ is $$ a_n = \frac{1}{{n!}}\left[ {\frac{{\mathrm{d}^n }}{{\mathrm{d}t^n }}\exp \left( {\sum\limits_{k = 1}^n {B_{k + 1}\! \left( {\tfrac{3}{2} - n} \right)\frac{{t^k }}{{k(k + 1)}}} } \right)} \right]_{t = 0} . $$ I omit the proof.

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} \mathrm{e}^{ - nt} \mathrm{d}t} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {\mathrm{e}^{ - xp(t)} q(t) \mathrm{d}t} , $$ with $p(t) = - t\mathrm{e}^{ - t}$ and $q(t)=\mathrm{e}^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim \mathrm{e}^{z} \sqrt {2\pi z} \sum\limits_{k = 0}^\infty {\frac{{a_k }}{{z^k }}} = \mathrm{e}^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/\mathrm{e}$ and $$ a_k = \frac{1}{{2^k k!}}\left[ {\frac{{\mathrm{d}^{2k} }}{{\mathrm{d}t^{2k} }}\left( {\mathrm{e}^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)\mathrm{e}^{ - t} }}} \right)^{k + 1/2} } \right)} \right]_{t = 0} . $$ An alternative expression for the coefficients involving the Bernoulli polynomials $B_j(x)$ is $$ a_k = \frac{1}{k!}\left[ \frac{\mathrm{d}^k }{\mathrm{d}t^k}\exp \left( \sum\limits_{j = 1}^k B_{j + 1}\! \left( \tfrac{3}{2} - k \right)\frac{t^j}{j(j + 1)} \right) \right]_{t = 0} . $$ I omit the proof.

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} e^{ - nt} dt} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {e^{ - xp(t)} q(t) dt} , $$ with $p(t) = - te^{ - t}$ and $q(t)=e^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim e^{z} \sqrt {2\pi z} \sum\limits_{n = 0}^\infty {\frac{{a_n }}{{z^n }}} = e^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/e$ and $$ a_n = \frac{1}{{2^n n!}}\left[ {\frac{{d^{2n} }}{{dt^{2n} }}\left( {e^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)e^{ - t} }}} \right)^{n + 1/2} } \right)} \right]_{t = 0} . $$ An alternative expression for the coefficients involving the Bernoulli polynomials $B_k(x)$ is $$ a_n = \frac{1}{{n!}}\left[ {\frac{{d^n }}{{dt^n }}\exp \left( {\sum\limits_{k = 1}^n {B_{k + 1}\! \left( {\tfrac{3}{2} - n} \right)\frac{{t^k }}{{k(k + 1)}}} } \right)} \right]_{t = 0} . $$ I omit the proof.

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} \mathrm{e}^{ - nt} \mathrm{d}t} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {\mathrm{e}^{ - xp(t)} q(t) \mathrm{d}t} , $$ with $p(t) = - t\mathrm{e}^{ - t}$ and $q(t)=\mathrm{e}^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim \mathrm{e}^{z} \sqrt {2\pi z} \sum\limits_{n = 0}^\infty {\frac{{a_n }}{{z^n }}} = \mathrm{e}^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/\mathrm{e}$ and $$ a_n = \frac{1}{{2^n n!}}\left[ {\frac{{\mathrm{d}^{2n} }}{{\mathrm{d}t^{2n} }}\left( {\mathrm{e}^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)\mathrm{e}^{ - t} }}} \right)^{n + 1/2} } \right)} \right]_{t = 0} . $$ An alternative expression for the coefficients involving the Bernoulli polynomials $B_k(x)$ is $$ a_n = \frac{1}{{n!}}\left[ {\frac{{\mathrm{d}^n }}{{\mathrm{d}t^n }}\exp \left( {\sum\limits_{k = 1}^n {B_{k + 1}\! \left( {\tfrac{3}{2} - n} \right)\frac{{t^k }}{{k(k + 1)}}} } \right)} \right]_{t = 0} . $$ I omit the proof.

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} e^{ - nt} dt} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {e^{ - xp(t)} q(t) dt} , $$ with $p(t) = - te^{ - t}$ and $q(t)=e^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim e^{z} \sqrt {2\pi z} \sum\limits_{n = 0}^\infty {\frac{{a_n }}{{z^n }}} = e^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/e$ and $$ a_n = \frac{1}{{2^n n!}}\left[ {\frac{{d^{2n} }}{{dt^{2n} }}\left( {e^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)e^{ - t} }}} \right)^{n + 1/2} } \right)} \right]_{t = 0} . $$

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} e^{ - nt} dt} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {e^{ - xp(t)} q(t) dt} , $$ with $p(t) = - te^{ - t}$ and $q(t)=e^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim e^{z} \sqrt {2\pi z} \sum\limits_{n = 0}^\infty {\frac{{a_n }}{{z^n }}} = e^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/e$ and $$ a_n = \frac{1}{{2^n n!}}\left[ {\frac{{d^{2n} }}{{dt^{2n} }}\left( {e^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)e^{ - t} }}} \right)^{n + 1/2} } \right)} \right]_{t = 0} . $$ An alternative expression for the coefficients involving the Bernoulli polynomials $B_k(x)$ is $$ a_n = \frac{1}{{n!}}\left[ {\frac{{d^n }}{{dt^n }}\exp \left( {\sum\limits_{k = 1}^n {B_{k + 1}\! \left( {\tfrac{3}{2} - n} \right)\frac{{t^k }}{{k(k + 1)}}} } \right)} \right]_{t = 0} . $$ I omit the proof.