I want to study the convergence of the function given below.
$$ \sum_{n = 1}^{\infty} e^{-\Gamma^2 n^2} $$
The $n$ are integers!
Here, if we check numerically, the function converges based on $\Gamma^2$ I believe. Is there a way to define a value of Number of summation elements $ N $ where the summation converges numerically?
The infinite sum given by mathematica is the following,
$$ \sum_{n = 1}^{\infty} e^{-\Gamma^2 n^2} = EllipticTheta[3, 0, e^{-\Gamma^2}] $$
I want an expression for the number of elements to sum where it converges numerically.
$$ \sum_{n = 1}^{\infty} e^{-\Gamma^2 n^2} \sim \sum_{n = 1}^{N} e^{-\Gamma^2 n^2} $$
Therefore, I need,
$$ N = f(\Gamma^2) $$
Is it possible?
For example, for a $\Gamma^2 = 0.17546$, the value of $N$ I observed on MATLAB is 7.
If you want an accuracy of $\varepsilon > 0$ (e.g. $\varepsilon=10^{-4}$, as mentioned in comments), it means that you need to find such $N$ that $$ \left|\sum_{n=1}^{\infty} e^{-\Gamma^2 n^2} - \sum_{n=1}^{N} e^{-\Gamma^2 n^2}\right| = \left|\sum_{n=N+1}^{\infty} e^{-\Gamma^2 n^2}\right| = \sum_{n=N+1}^{\infty} e^{-\Gamma^2 n^2} < \varepsilon $$ Observe that $$ \sum_{n=N+1}^{\infty} e^{-\Gamma^2 n^2} = \sum_{n=N+1}^{\infty} e^{-\Gamma^2(n^2 - n)} e^{-\Gamma^2 n} \leq \sum_{n=N+1}^{\infty} e^{-\Gamma^2(N^2 + N)} e^{-\Gamma^2 n} = e^{-\Gamma^2(N^2 + N)} \sum_{n=N+1}^{\infty}e^{-\Gamma^2 n} $$ The inequality is due to $n^2 - n = n(n-1) \geq (N+1)N$ for $n\geq N+1$. By the geometric series sum formula, $$ \sum_{n=N+1}^{\infty}e^{-\Gamma^2 n} = \frac{e^{-\Gamma^2 (N+1)}}{1 - e^{-\Gamma^2}}, $$ so $$ \sum_{n=N+1}^{\infty} e^{-\Gamma^2 n^2} \leq e^{-\Gamma^2 (N^2 + N)} \cdot \frac{e^{-\Gamma^2 (N+1)}}{1 - e^{-\Gamma^2}} = \frac{e^{-\Gamma^2(N+1)^2}}{1 - e^{-\Gamma^2}} $$ and it's needed to solve $\frac{e^{-\Gamma^2(N+1)^2}}{1 - e^{-\Gamma^2}} < \varepsilon$: $$ \frac{e^{-\Gamma^2(N+1)^2}}{1 - e^{-\Gamma^2}} < \varepsilon \Leftrightarrow e^{-\Gamma^2(N+1)^2} < \varepsilon \cdot \left(1 - e^{-\Gamma^2}\right) \Leftrightarrow \\ \Leftrightarrow \Gamma^2 (N+1)^2 > -\ln \varepsilon - \ln \left(1 - e^{-\Gamma^2}\right) \Leftrightarrow N + 1 > \frac{1}{\Gamma}\sqrt{- \ln \varepsilon - \ln \left(1 - e^{-\Gamma^2}\right)} $$ So, $N = f(\Gamma) = \left\lceil\frac{1}{\Gamma}\sqrt{- \ln \varepsilon - \ln \left(1 - e^{-\Gamma^2}\right)} - 1 \right\rceil$. Here is also a Desmos link for testing this bound.