Connection between $J_k= \sum_{n=1}^k e^{-n}=\frac{1-e^{-k}}{e-1} $ and $ f(x)=\sum_{n=1}^\infty e^{-n^x} ?$

Question

Consider the geometric series$$J_k= \sum_{n=1}^k e^{-n}=\frac{1-e^{-k}}{e-1} $$

I'm wondering if this has any connection with:

$$ f(x)=\sum_{n=1}^\infty e^{-n^x}. $$

$J_k$ can be interpreted as adding up y-values on the curve $y=e^{\frac{1}{\log x}}$ for $x=e^{-1/n}$ in the same way that $H_k$ can be interpreted as adding up y-values on the curve $y=1/x$ for $x=n\ge1.$

Now the Harmonic numbers, $H_k$ are closely related to the Riemann zeta function, $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}.$

Might there be a close relationship between $J_k$ and $f(x)?$

Please help me understand why or why not?

Answer 1

First a few observations:

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(1) $\quad\sum\limits_{n=1}^\infty e^{-n}=\frac{1}{e-1}$

(2) $\quad\sum\limits_{n=1}^\infty \left(e^{-n}\right)^s=\frac{1}{e^s-1}$

(3) $\quad\sum\limits_{n=1}^\infty e^{-n^2}=\frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)-1\right)$

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Now consider the function $f(x)$ defined in formula (3) below and the related function $F(s)$ defined in formula (4) below.

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(3) $\quad f(x)=\sum\limits_{n\le x} a(n)$

(4) $\quad F(s)=\underset{N\to\infty}{\text{lim}}\ \sum\limits_{n=1}^N\frac{a(n)}{n^s}$

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Setting $a(n)=1$ yields the Riemann zeta function $\zeta(s)$ as illustrated in formula (5) below, and setting $a(n)=e^{-n}$ yields the PolyLog function $\text{Li}_s\left(\frac{1}{e}\right)$ as illustrated in formula (6) below.

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(5) $\quad\underset{N\to\infty}{\text{lim}}\ \sum\limits_{n=1}^N \frac{1}{n^s}=\zeta(s),\quad \Re(s)>1$

(6) $\quad\underset{N\to\infty}{\text{lim}}\ \sum\limits_{n=1}^N\frac{e^{-n}}{n^s}=\text{Li}_s\left(\frac{1}{e}\right)$

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Note formula (5) above diverges at $s=1$ as $N\to\infty$ (since the Harmonic series diverges), but setting $s=1$ in formula (6) above yields the following.

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(7) $\quad\underset{N\to\infty}{\text{lim}}\ \sum\limits_{n=1}^N \frac{e^{-n}}{n}=\text{Li}_1\left(\frac{1}{e}\right)=-\log \left(1-\frac{1}{e}\right)$

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The partial sums associated with formulas (5) and (6) at $s=1$ are as follows where $\Phi()$ is the Lerch transcendent.

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(8) $\quad\sum\limits_{n=1}^N \frac{1}{n}=H_N$

(9) $\quad\sum\limits_{n=1}^N \frac{e^{-n}}{n}=-\log \left(1-\frac{1}{e}\right)-e^{-N-1} \Phi \left(\frac{1}{e},1,N+1\right)$