Return to Answer

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > 1+\frac{1}{e}f \tag{2} $$ Since $f(0) = 0$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=0$ is given by $g(x) = e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e+1}-e > 2e^{x/e} , \qquad x>4\tag{3} $$ ($x>4$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{4} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq e^x - 1\tag{5} $$

Overall, for $x>4$, $$ 2e^{x/e} \leq f(x) \leq e^x - 1 \tag{6} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

Here is a plot of $\log f(x)$, $\log(2e^x - 1)$ (instead of $\log(e^x - 1)$, due to a mistake -- this changes almost nothing since it's still $x+O(1)$), and $\log(2e^{x/e})$, for $x$ ranging from $1$ to $50$.

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

The main idea: bounding $f$ via differential partial equation.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > 1+\frac{1}{e}f \tag{2} $$ Since $f(0) = 0$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=0$ is given by $g(x) = e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e+1}-e > 2e^{x/e} , \qquad x>4\tag{3} $$ ($x>4$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{4} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq e^x - 1\tag{5} $$

Overall, for $x>4$, $$ 2e^{x/e} \leq f(x) \leq e^x - 1 \tag{6} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

Further: Improving (slightly) on the lower bound on $\log f$ by the low-order terms, and improving on the constant in the main asymptotics of the upper bound of $\log f$.

I will show $$ h(x) \leq f(x) \leq g(x) \tag{7} $$ where $$ \begin{align} \log h(x) &= \frac{1}{e}x + 4 - \log\frac{32}{3} + o(1) \tag{8}\\ \log g(x) &= \frac{256}{625}x + O(1) \tag{9} \end{align} $$ (note that $\frac{256}{625} \approx \frac{1}{e}+0.04$). Moreover, this can be improved by the same method, pushing to more accurate estimates, but this will get uglier. (One can also push the Taylor expansion above further, based on (12) and (13). I stopped at $o(1)$).

The observation is that for the upper and lower bound, we bounded uniformly the coefficients by $$ \forall n \geq 1\, \qquad \frac{1}{n^n} \leq \frac{1}{\left(1+\frac{1}{n}\right)^{n}} \cdot \frac{1}{n^n} \leq \frac{1}{e}\cdot \frac{1}{n^n} $$ to obtain the two corresponding differential equations. We can do better by handling the first few terms more tightly. Namely, we have $$ \left(1+\frac{1}{n}\right)^n = \begin{cases} \frac{1}{2} & n=1\\ \frac{4}{9} & n=2\\ \frac{27}{64} & n=3\\ \frac{256}{625} & n=4 \end{cases} $$ (and, of course, $\left(1+\frac{1}{n}\right)^n$ is decreasing to $1/e$). Thus, we can leverage this and solve instead the following two differential equations for $h$ and $g$: \begin{align} h'(x) &= 1 + \left(\frac{1}{2} - \frac{1}{e}\right) x + \left(\frac{4}{9} - \frac{1}{e}\right) \frac{x^2}{4} + \left( \frac{27}{64} - \frac{1}{e}\right) \frac{x^3}{27} + \frac{1}{e}h(x)\tag{10}\\ g'(x) &= 1 + \left(\frac{1}{2} - \frac{256}{625}\right) x + \left(\frac{4}{9} - \frac{256}{625}\right) \frac{x^2}{4} + \left( \frac{27}{64} - \frac{256}{625}\right) \frac{x^3}{27} + \frac{256}{625}g(x)\tag{11} \end{align} subject to $h(0)=g(0)=0$. Solving those gives a nasty expression, \begin{align} h(x) &= \frac{3}{32} e^{4 + \frac{1}{e}x} + \left(\frac{1}{27} - \frac{e}{64}\right) x^3 + \left(\frac{1}{4} - \frac{3 e^2}{64}\right) x^2 + \left(1 - \frac{3e^3}{32}\right) x -\frac{3e^4}{32} \tag{12} \\ g(x) &= \frac{457763671875}{137438953472}e^{\frac{256}{625}x} - \frac{491}{442368}x^3 - \frac{123299}{4194304}x^2 - \frac{195550963}{536870912}x -\frac{457763671875}{137438953472} \tag{13} \\ \end{align} leading to the claimed (8) and (9).

Below, a plot illustrating those approximations:

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > \frac{1}{e}f \tag{2} $$ Since $f(0) = 1$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=1$ is given by $g(x) = e^{x/e}+e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e}+e^{x/e+1}-e > 2e^{x/e} , \qquad x>2\tag{3} $$ ($x>2$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{2} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq 2e^x - 1\tag{3} $$

Overall, for $x>2$, $$ 2e^{x/e} \leq f(x) \leq 2e^x - 1 \tag{4} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

Here is a plot of $\log f(x)$, $\log(2e^x - 1)$, and $\log(2e^{x/e})$, for $x$ ranging from $1$ to $50$.

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > 1+\frac{1}{e}f \tag{2} $$ Since $f(0) = 0$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=0$ is given by $g(x) = e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e+1}-e > 2e^{x/e} , \qquad x>4\tag{3} $$ ($x>4$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{4} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq e^x - 1\tag{5} $$

Overall, for $x>4$, $$ 2e^{x/e} \leq f(x) \leq e^x - 1 \tag{6} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

Here is a plot of $\log f(x)$, $\log(2e^x - 1)$ (instead of $\log(e^x - 1)$, due to a mistake -- this changes almost nothing since it's still $x+O(1)$), and $\log(2e^{x/e})$, for $x$ ranging from $1$ to $50$.

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > \frac{1}{e}f \tag{2} $$ Since $f(0) = 1$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=1$ is given by $g(x) = e^{x/e}+e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e}+e^{x/e+1}-e > 2e^{x/e} , \qquad x>2\tag{3} $$ ($x>2$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{2} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq 2e^x - 1\tag{3} $$

Overall, for $x>2$, $$ 2e^{x/e} \leq f(x) \leq 2e^x - 1 \tag{4} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

Here is a plot of $\log f(x)$, $\log(2e^x - 1)$. and $\log(2e^{x/e})$, for $x$ ranging from $1$ to $50$.

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > \frac{1}{e}f \tag{2} $$ Since $f(0) = 1$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=1$ is given by $g(x) = e^{x/e}+e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e}+e^{x/e+1}-e > 2e^{x/e} , \qquad x>2\tag{3} $$ ($x>2$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{2} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq 2e^x - 1\tag{3} $$

Overall, for $x>2$, $$ 2e^{x/e} \leq f(x) \leq 2e^x - 1 \tag{4} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

Here is a plot of $\log f(x)$, $\log(2e^x - 1)$, and $\log(2e^{x/e})$, for $x$ ranging from $1$ to $50$.