# Interpolating $G(1)=\sum_{a=1}^{\infty} \frac{1}{a^{a}}$, $G(2) = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{1}{(ab)^{ab}}$ on $\mathbb{C}$

Given that: $$G(1) =\sum_{a=1}^{\infty} \frac{1}{a^{a}}$$

(this is just the Sophomore's dream series, but the rest are not)

$$G(2) = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{1}{(ab)^{ab}}$$

$$G(3) = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(abc)^{abc}}$$

I'd like to interpolate $G(z)$ for $z\in\mathbb{C}$ (so that the result is analytic or meromorphic) given the above sequence. I can probably compute (with difficulty) $G(n)$ for many $n$.

• Does the sequence above specify a unique $G(z)$?
• Are there analytic tricks which would make finding such an interpolation easy? (the less I have to compute here, the better) *
-
Perhaps you are asking for $G$ to be analytic (or at any rate meromorphic). If so, please edit that into the body of the question. – Gerry Myerson Jan 11 at 4:35

Just playing around to try to get another form for $G(2)$.

$\begin{array}\\ G(2) &= \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{1}{(ab)^{ab}}\\ &= \sum_{n=1}^{\infty} \sum_{d|n} \frac{1}{n^{n}}\\ &= \sum_{n=1}^{\infty}\frac{1}{n^{n}} \sum_{d|n} 1 \\ &= \sum_{n=1}^{\infty}\frac{d(n)}{n^{n}} \\ \end{array}$

Don't see where to go from here, so I'll leave it at this.

-
More generally, $$G(m) = \sum_{n=1}^\infty \dfrac{\tau_m(n)}{n^n}$$ where $\tau_m(n)$ is the $m$'th Piltz function, the number of ordered $m$-tuples of positive integers whose product is $n$. – Robert Israel Jan 15 at 22:47
Thanks. Hadn't heard about that series of functions. – marty cohen Jan 15 at 22:54
Neither had I. But see e.g. OEIS sequence A007425. – Robert Israel Jan 15 at 22:56

For any sequence $a_n$ of complex numbers, there are uncountably many entire functions $G(z)$ such that $G(n) = a_n$ for all positive integers $n$. One way to obtain one of them is $G(z) = \sum_{n=1}^\infty G_n(z)$, where \eqalign{G_1(z) &= a_1\cr G_n(z) &= \left(a_n - \sum_{j=1}^{n-1} G_j(n)\right) \left(\frac{z}{n}\right)^{k_n} \prod_{j=1}^{n-1} \dfrac{z-j}{n-j}\ \text{for}\ n > 1} where $k_n$ is taken large enough so that $|G_n(z)| < 2^{-n}$ for $|z| < n/2$.

Of course you can add functions such as $\sin(\pi z)$ to solutions to obtain others.

-