Convergence of a series of random elements

Question

Given the normally distribuited random variable $\nu(t)$ with $\mu=0$ and variance $\sigma$, I have to find if the series: $$G(\sigma)=\sum_{k=1}^{\infty}\frac{1}{\exp\left(\nu(k)\right)}$$ where $\nu(k)$ is the random value of the $\nu(t)$ for $t=1,2,...$, is convergent and if it is, how can be evaluated the result of the sum. Thanks

Answer 1

As long as the parameters of the distribution of $\nu(t)$ do not depend on $t$, we can simply refer to this variable as $\nu$. Thus, $\frac{1}{\exp(\nu)}$ does not converge to anything, so the infinite sum of it goes to infinity.