Prove that the number: $$\sum_{n=-\infty}^{\infty} \frac{1}{2^{n^2}}$$ is transcendental.
I don't have a direct proof but a round one. The series can be expressed in terms of $\vartheta_3$ which is one of the theta Jacobi functions. More specifically,
$$\sum_{n=-\infty}^{\infty} \frac{1}{2^{n^2}}=\vartheta_{3} \left (\frac{1}{2} \right)$$
From the theory of the Jacobi function we know that for $|q|<1$ the series converges and the number is transcedental.
Any other direct proof?
A very similar question has been answered on MathOverflow. Basically, it's known that
$$\sum_{n=1}^\infty q^{n^2}$$
is transcendental if $q$ is algebraic and $0<|q|<1$. Thus, taking $q=\frac12$, we see
$$\alpha=\sum_{n=1}^\infty \left(\frac12\right)^{n^2}=\sum_{n=1}^\infty \frac1{2^{n^2}}$$
is transcendental, and thus, your constant is \begin{align} \sum_{n=-\infty}^\infty \frac1{2^{n^2}}&=\left(\sum_{n=-\infty}^{-1}\frac1{2^{n^2}}\right)+1+\left(\sum_{n=1}^\infty \frac1{2^{n^2}}\right)\\ &=\left(\sum_{n=1}^{\infty}\frac1{2^{n^2}}\right)+1+\alpha\\ &=2\alpha+1 \end{align}
so yes, your number is transcendental.