Show that $\sum\limits_{j,k=2}^\infty\frac{1}{j^k}$ converges and calculate the limit of the series

Formally,

$$ \sum_{j = 2}^{\infty} \left( \sum_{k = 2}^\infty \frac{1}{j^k} \right) = \sum_{j = 2}^{\infty} \left( \frac{\frac{1}{j^2}}{1 - \frac{1}{j}} \right) = \sum_{j = 2}^{\infty} \left( \frac{1}{j( j - 1)} \right) = \sum_{j = 2}^{\infty} \left( \frac{1}{j - 1} - \frac{1}{j} \right) = 1. $$

and since all terms $\frac{1}{j^k}$ are positive the computations are justified.

Your computations are correct, I would just use a more understandable formula for the sum of $\frac{1}{j^k}$ on $k$. (but once again, you computation is totally fine !)

NOTE: I am not sure what you are referring to can be called Cauchy's product rule (or at least it is not the one that I know of). See https://en.wikipedia.org/wiki/Cauchy_product for more precisions on that.