An upper bound for $1+\sum_{k=1}^\infty (\frac{a}{k})^k$, better than $\exp(a)$

Question

For $a>0$ is the result of $1+\sum_{k=1}^\infty (\frac{a}{k})^k$ known, or can it be upper-bounded? There is an obvious upper bound of $exp(a)$ resulting from $k^k\ge k!$ and hence: $$1+\sum_{k=1}^\infty (\frac{a}{k})^k \le \sum_{k=0}^\infty \frac{a^k}{k!} = \exp(a) $$ But can this bound be made tighter?

Answer 1

This bound derives from asymptotic analysis and works best for large $a,$ although it works for any $a \ge 0.$ For convenience, I'm dropping the +1. I'll show $$g(a):=\sum_{k=1}^\infty \big(\frac{a}{k}\big)^k < \sqrt{2 \pi x} \exp(x) \text{ where } x=a/e.$$ From Stirling's formula for the factorial, $$ \frac{k!}{k^k} = e^{-k}\sqrt{2 \pi k} \ \Big(1+\frac{1}{12k} + \frac{1}{288k^2} - ...\Big)$$ The important thing is that the terms eventually get negative signs, so if we want a bound, we can't go any further than the ones shown. Now it's an asymptotic series, so for small $k,$ we don't expect it to be terribly accurate. However, numerically it can be seen that $$ k^{-k} \le \sqrt{2 \pi k} \ \frac{ e^{-k} }{k!}\Big(1 + \frac{1}{11.84k} \Big) \text{ where } k=1, 2,...$$ Thus $$g(a) < \sqrt{2 \pi} \sum_{k=1}^\infty \frac{x^k}{k!}\sqrt{k} \Big(1+ \frac{1}{11.84k} \Big)\quad,\quad x=a/e. $$ It can be shown (MSE 3309513) that $$ \sum_{k=1}^\infty \frac{x^k}{k!} k^\epsilon \sim x^\epsilon e^x\Big(1 - \frac{\epsilon(1-\epsilon)}{2x}+...\Big) $$ Use $\epsilon=1/2$, and collect the terms properly to get $$g(a) < \sqrt{2 \pi x}e^x\Big( \big(1 - \frac{1}{8x} \big) + \frac{1}{11.84 x} \Big) .$$ The coefficient before the 1/x term is -0.0405, the important thing being it is negative. Dropping the 1/x term means the inequality will still be satisfied.

I've checked this for all $a<100,$ and of course the large $a$ case reverts to asymptotics, of which this is the first term. For $a$ as small as 14, the error between 'truth' and the bound is about 1%. For numerical checking, $g(14)=971.09$ and the approximation $\tilde{g}(14)=981.203.$

Answer 2

Define $$f(a)=1+\sum_{k=1}^{\infty}\left({a\over k}\right)^k$$therefore $$f'(a){=\sum_{k=1}^{\infty}{ka^{k-1}\over k^k}\\=\sum_{k=1}^{\infty}{ka^{k-1}\over k^k}\\=\sum_{k=1}^{\infty}{a^{k-1}\over k^{k-1}}\\=\sum_{k=0}^{\infty}{a^{k}\over (k+1)^{k}}\\=1+\sum_{k=1}^{\infty}{a^{k}\over k^{k}}({k\over k+1})^k}$$Since the sequence $(1+{1\over k})^k$ is strictly increasing, then $({k\over k+1})^k$ becomes strictly decreasing and $\le {1\over 2}$ (which can be visually checked in https://www.desmos.com/calculator). Hence $$f'(a)\le 1+{1\over 2}\exp(a)$$and by integrating we obtain $$f(a)\le{1\over 2}+a+{1\over 2}\exp(a)$$or

$$1+\sum_{k=1}^{\infty}\left({a\over k}\right)^k\le {1\over 2}+a+{1\over 2}\exp(a)$$

Answer 3

First, let us obtain the integral representation of the series. Note that $$ \frac{1}{k^k} = \frac{1}{(k-1)!}\int_0^\infty e^{-kx}x^{k-1}dx, \quad k\in \mathbb{N}_{\ge 1}.$$ We have \begin{align} 1 + \sum_{k=1}^\infty \left(\frac{a}{k}\right)^k &= 1 + \sum_{k=1}^\infty a^k \frac{1}{(k-1)!}\int_0^\infty e^{-kx}x^{k-1}dx\\ &= 1 + \int_0^\infty ae^{-x} \sum_{k=1}^\infty \frac{(ae^{-x}x)^{k-1}}{(k-1)!} dx\\ &= 1 + \int_0^\infty ae^{-x} e^{ae^{-x}x} dx\\ &= 1 + \int_0^1 ay^{-ay} dy\\ &= 1 + \int_0^1 ae^{-ay\ln y} dy. \end{align} Note that $-y\ln y \le \frac{1}{e} - \frac{5}{2e}(y- \frac{1}{e})^2$ for $0 < y < 1$. An upper bound is given by \begin{align} 1 + \int_0^1 ae^{-ay\ln y} dy &\le 1 + e^{a/e}a\int_0^1 e^{-a\frac{5}{2e}(y- \frac{1}{e})^2} dy\\ &= 1 + e^{a/e}a \frac{\sqrt{\pi}}{2\sqrt{q}} \Big[\mathrm{erf}(\sqrt{q}\, e^{-1}) - \mathrm{erf}(-\sqrt{q} + \sqrt{q}\, e^{-1})\Big]\qquad (1) \end{align} where $q = \frac{5a}{2e}$. The following figure shows the upper bound (1) and the upper bound $1+\sqrt{2\pi a/e}e^{a/e}$ derived by skbmoore. It can be seen that when $a$ is small, the upper bound (1) is better than $1+\sqrt{2\pi a/e}e^{a/e}$. However, when $a > 4.5$, the upper bound (1) is worse.

I think that we can obtain better (or simpler) upper bounds by analyzing $1 + \int_0^1 ae^{-ay\ln y} dy$.