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