Answer
converges
Work Step by Step
Since for $1 \leq \theta \lt\infty ;~ 0 \leq \dfrac{1}{1+e^{\theta}} \leq \dfrac{1}{e^{\theta}}$
We are given that: $\int_{0}^{\infty} \dfrac{d \theta}{1+e^{\theta}}$
Now, $\int_0^{\theta} \dfrac{1}{e^{\theta}} = \lim\limits_{a \to \infty}[-e^{\theta}]_0^b \\ = \lim\limits_{a \to \infty}[-e^{-b}+1]$
This means that the integral $\int_0^{\theta} \dfrac{1}{e^{\theta}}$ converges and the integral $\int_{0}^{\infty} \dfrac{d \theta}{1+e^{\theta}} $ converges by the direct comparison test.