About the sums $\sum_{n=1}^\infty x^{n^2}$ and $\sum_{n=1}^\infty \frac{x^n}{1+x^{2n}}$ Despite all my efforts trying to crack these, i haven't been able to do so. An approach that i've tried gives me somewhat of an asymptotic approximation, but still fails to produce the values near x=0.
The method i've tried is to express the sum as an integral and then integrate it to obtain it's value but apparently there is something that I can't do or something wrong with it.
If anybody can suggest any ideas or (ideally) solve the problem I'd be very grateful. Thanks in advance.
I'll try to show here what I think i did wrong, maybe then you'll be able to point out where i fail at.
Since the two sums are related, i'll only try to find one of them, the second one. Since:
$$ \int_a^b f(x)dx = \lim_{n\to \infty} \sum_{k=1}^n f(a+\frac{b-a}{n}k)\cdot \frac{b-a}{n}$$
I let each term of the sum be $\frac{c^k}{1+c^{2k}}$ so, if $x=a+ \frac{b-a}{n}k$ then $f(x)=\frac{n}{b-a}\cdot \frac{c^{\frac{x-a}{b-a}n}}{1+c^{\frac{x-a}{b-a}2n}}$ therefore: $$\sum_{k=1}^\infty \frac{c^k}{1+c^{2k}} = \lim_{n\to \infty} \int_a^b \frac{n}{b-a}\cdot \frac{c^{\frac{x-a}{b-a}n}}{1+c^{\frac{x-a}{b-a}2n}}dx$$ and integrating between a and b I obtain (since c<1 and n tends to infinity):
$$\sum_{k=1}^\infty \frac{c^k}{1+c^{2k}}=\frac{\pi}{4log(\frac{1}{c})}$$ Therefore $$\sum_{n=1}^\infty x^{n^2} = \frac{\sqrt{1-\frac{\pi}{log(x)}}-1}{2}$$ and as x tends to 1- : $$\sum_{n=1}^\infty x^{n^2} = \frac{\sqrt{1+\frac{\pi}{1-x}}-1}{2}$$
But STILL this only works as an asymptotic function. Obviously there is something wrong with this reasoning. Hope you guys can help me. Thanks a lot again.
*Edit: In the photo i've posted you can see the difference between the graph of the functions.
 A: Let me first summarize some essential comments before proceeding to an answer.


*

*As JJacquelin has hinted,
$$\sum_{n=1}^\infty x^{n^2} = \frac{\theta_3(0,x)-1}{2}$$
where $\theta_3(0,x)$ is a
Jacobi thetanull
$$\theta_3(0,x) = \sum_{n=-\infty}^\infty x^{n^2}\qquad(|x| < 1)$$
Cf. DLMF ch. 20 Theta functions and
OEIS A010052.

*As you have noted,
$$\sum_{n=1}^\infty x^{n^2} =
   \frac{\sqrt{1+4 \sum_{n=1}^\infty \frac{x^n}{1+x^{2n}}}-1}{2}$$
Consequently,
$$\begin{align}
   \theta_3(0,x)^2 &= 1 + 4\sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} \\
   \sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} &= \frac{\theta_3(0,x)^2-1}{4}
   \end{align}$$
Cf. OEIS A002654 and
OEIS A004018
where you can find other representations as well, e.g.
$$\sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} =
   \sum_{k=0}^\infty (-1)^k\frac{x^{2k-1}}{1-x^{2k-1}}$$
which can be obtained by expanding the summand to a power series,
changing the nesting order of the summation, and unexpanding
the new inner series.

*As Dr. MV has pointed out,
your $f$ is not allowed to depend on the step count $n$ whilst
approximating the integral.


I'd like to add that the modular properties of $\theta_3$ allow you to convert between
values at small $|x|$ and values at $|x|$ near $1$:
$$\theta_3\!\left(0,\exp(-\pi/y)\right)
    = \sqrt{y}\,\theta_3\!\left(0,\exp(-\pi y)\right)
    \qquad(\Re y > 0)\tag{*}$$
Note that the associated lattice parameter (period ratio) is
$\tau=\mathrm{i}y$ here.
To answer your question:
For computation of $\theta_3(0,x)$ at nonnegative $x$, the
DLMF notes on computation work quite well.
If $|x|\leq\exp(-\pi)$, the series for $\theta_3(0,x)$ converges very quickly.
Otherwise, you can use the transform $(*)$ which recurs to a new value of
$x$ less than $\exp(-\pi)$.
For negative $x$, compute $\theta_3(0,x)$ as $\theta_4(0,-x)$ in a similar
manner, but note that the modular transformation then swaps $\theta_4$ with
$\theta_2$:
$$\theta_4\!\left(0,\exp(-\pi/y)\right)
    = \sqrt{y}\,\theta_2\!\left(0,\exp(-\pi y)\right)
    \qquad(\Re y > 0)$$
For complex nonzero $x$, the above approach must be slightly refined.
Set the lattice parameter $\tau=\frac{\log x}{\pi\mathrm{i}}$.
Since $|x|<1$, we have $\Im\tau>0$.
Now either recursively compute the true two-variable version $\theta_3(z\mid\tau)$
(initially for $z=0$) or the triple
$$T(\tau) = \left(\theta_2(0\mid\tau),\theta_3(0\mid\tau),\theta_4(0\mid\tau)\right)$$
I'll describe the latter approach.
We have the modular symmetries
$$\begin{align}
T(\tau+1) &= \left(\sqrt{\mathrm{i}}\,\theta_2(0\mid\tau),
\theta_4(0\mid\tau),\theta_3(0\mid\tau)\right) \tag{T}\\
T(-1/\tau) &= \sqrt{-\mathrm{i}\tau}
\left(\theta_4(0\mid\tau),\theta_3(0\mid\tau),\theta_2(0\mid\tau)\right)
\tag{J}
\end{align}$$
If $\Im\tau$ is greater than some threshold, say $\frac{1}{2}$,
then the associated $|x|$ is very small. In that case, use the series
representations for the entries of $T(\tau)$ with suitable truncation.
Otherwise, use (T) to reduce the real part of $\tau$ such that $|\Re\tau|\leq\frac{1}{2}$. Since $\Im\tau$ is bounded here, we also have
$|\tau|<1$, in fact, with the above threshold $|\tau|\leq\frac{1}{\sqrt{2}}$.
Now use (J) to turn to some $|\tau|>1$; the above-proposed threshold gets us
$|\tau|\geq\sqrt{2}$; in particular, $\Im\tau$ at least doubles in that step.
Repeating those recursions gets you quickly to a state where $\Im\tau$
is large enough that you can use the truncated series.
