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?


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.


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 $x$.

  • $\begingroup$ Thanks, edited the typos $\endgroup$ – Mike Battaglia Oct 15 '18 at 18:18

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