Define the null theta funtion
$$ f(q) := \sum_{n=-\infty}^\infty q^{n^2} = 1 +
2\sum_{n=1}^{\infty} q^{n^2} \tag{1} $$
where $\,|q|<1\,$ is needed for convergence. Assume that further
$\,0<q<1.\,$ This implies that truncating the infinite sum will
result in increasing lower bounds such as
$$ 1 < 1 + 2q < 1 + 2q + 2q^4 < \cdots < f(q). \tag{2} $$
However, comparing it to a geometric series gives an upper bound
$$ f(q) < 1 + 2\sum_{n=1}^\infty q^n = 1 + \frac{2q}{1-q} = \frac{1+q}{1-q} \tag{3} $$
although better upper bounds exist. For example,
$$ f(q) < 1 + 2\sum_{n=0}^\infty q^{3n+1} = 1+2q+\frac{2q^4}{1-q^3} \tag{3} $$ and so on. Thus,
$$ f(q) \!<\! \cdots \!<\! 1 \!+\! 2q \!+\! 2q^4 \!+\!
\frac{2q^9}{1\!-\!q^5} \!<\! 1 \!+\! 2q \!+\! \frac{2q^4}{1\!-\!q^3} \!<\! 1 \!+\! \frac{2q}{1\!-\!q}. \tag{4} $$