I am working on a problem and I want to take the derivative of the following (with respect to $x$) and evaluate at $x = 0$:
$\sum\limits_{i=1}^\infty\sum\limits_{j=1}^i \frac{j}{i} \, p_i \, q_j \, x^i \quad $ where $0\leq p_i,q_j\leq 1$ for all $i,j$.
So I know we can switch limits and derivatives when the series converges uniformly, but I am not sure how to see that here (I think $x$ must be in $(-1,1)$ here, but I don't think that is important since I want to evaluate at $x=0$). Thoughts?
This just looks like a power series in $x$ $$\sum_{i=1}^{\infty} \left(\sum_{j=1}^i jq_j\right)\dfrac{p_i}{i}x^i:=\sum_{i=1}^{\infty} a_ix^i$$ Your assumption on $p_i,q_j$ gives a bound on $a_i$ which should be enough to prove that the radius of convergence is at least $1$. The series is then uniformly convergent on $(-1,1)$ open.