Given is a set of random numbers generators, denounced by $U_k \sim Uni[0, \frac{1}{k^2}], \: k \geq 1$.
We will randomly choose a generator by rolling a Poisson dice with some constant $\lambda$ in the next fashion: Randomly pick from $I \sim Poi(\lambda)$ and add $1$. That is the index of the generator we will use.
We will poll a random number using said random generator, and call it $X$. Find the expected value $\mathbb{E}[X]$.
I'm unsure on how to approach this, I think that choosing a random generator is $\mathbb{E}[I] + 1$ but I feel like there is a conditional expectation here.
Update
Going off from the comments so far: $$\mathbb{E}[X] = \sum_{k = 1}^\infty \mathbb{E}[X \mid I = k] \cdot P(I = k) $$
Seeing how $X \mid I \sim Uni(0, \frac{1}{(I+1)^2})$, we get $P(X = k) = \frac{1}{(k + 1)^2}$.
So far this leads me to $$\begin{gather*} \begin{split} E[X] &= \sum_{k = 1}^\infty E[X \mid I = k] \cdot P(I = k) \\ &= \sum_{k = 1}^\infty \frac{1}{(k+1)^2} \cdot P(I = k) \\ &= \sum_{k = 1}^\infty \frac{1}{(k+1)^2} \cdot \frac{e^{- \lambda} \cdot \lambda^k}{k!} \\ &= e^{-\lambda} \cdot \sum_{k = 1}^\infty \frac{1}{(k + 1)^2} \cdot \frac{\lambda^k}{k!} \\ &= e^{-\lambda} \cdot \sum_{k = 1}^\infty \frac{\lambda^k}{(k + 1)(k+1)!} \end{split} \end{gather*} $$
Is this ok? Is there a way to continue developing?