Closed-form expression for infinite series related to a Gaussian Consider the following infinite series, where $x$ is indeterminate and $r$ is held constant:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^2} + \frac{x^3}{r^3} + ...$
It is relatively easy to see that the above, for $\frac{x}{r} < 1$, converges to
$\displaystyle \frac{1}{1-\frac{x}{r}}$
Now suppose we modify the above to this:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^{2^2}} + \frac{x^3}{r^{3^2}} + ...$
which we can rewrite as
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^4} + \frac{x^3}{r^9} + ...$
Does there then exist a well-known, closed-form expression for this series?
If not for general $r$, then perhaps for certain special values of $r$? For example, if we set $r=e$ above, then we get
$\displaystyle 1 + \frac{x}{e} + \frac{x^2}{e^4} + \frac{x^3}{e^9} + ...$
which we can rewrite as
$\displaystyle 1 + xe^{-1^2} + x^2e^{-2^2} + x^3e^{-3^2} + ...$
So that we can see that for x=1, this becomes a series of evenly spaced points on a Gaussian function.
Does there exist a closed-form expression for any of these?
 A: Interestingly, if we change the definition slightly, we get something related to the Jacobi theta function.
If we start with this series:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^4} + \frac{x^3}{r^9} + ...$
We can make the following substitutions:
$q = \frac{1}{r}$
$x = \exp(2\pi i z)$
to obtain
$\displaystyle 1 + q\exp(2\pi i z) + q^4\exp(4\pi i z) + q^9\exp(6\pi i z) + ...$
$ = \sum_0^\infty q^{n^2} \exp(2\pi i n z)$
If we simply change the bottom bound from $0$ to $\infty$, we get
$\theta_3(z;q) = \sum_{-\infty}^\infty q^{n^2} \exp(2\pi i n z)$
So it is easy to write the Jacobi theta function in terms of the function I described; it is probably possible to write it the other way as well.
A: Your original series is
$f(x, r)
=\sum_{n=0}^{\infty} \dfrac{x^n}{r^{(n^n)}}
$.
This is not the same as
$g(x, r)
=\sum_{n=0}^{\infty} \dfrac{x^n}{(r^n)^n}
=\sum_{n=0}^{\infty} \dfrac{x^n}{r^{n^2}}
$.
Also,
you went from
$\displaystyle 1 + \frac{x}{e} + \frac{x^2}{e^4} + \frac{x^3}{e^9} + ...
$
to
$\displaystyle 1 + x \cdot e^{-1^2} + x \cdot e^{-2^2} + x \cdot e^{-3^2} + ...
$,
somehow losing the
exponent of $x$.
