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

My approach:
We look if one of the iterated series converges absolutly.
$$\sum\limits_{j=2}^\infty\left(\sum\limits_{k=2}^\infty\left|\frac{1}{j^k}\right|\right)=\sum\limits_{j=2}^\infty\left(\frac{1}{1-\frac{1}{j}}-1-\frac{1}{j}\right)=\sum\limits_{j=2}^\infty\left(\frac{j}{j-1}-1-\frac{1}{j}\right)=\sum\limits_{j=2}^\infty\left(1+\frac{1}{j-1}-1-\frac{1}{j}\right)$$
$$=\sum\limits_{j=2}^\infty\left(\frac{1}{j-1}-\frac{1}{j}\right)=\sum\limits_{j=2}^\infty\left(\frac{j-(j-1)}{j(j-1)}\right)=\sum\limits_{j=2}^\infty\left(\frac{1}{j(j-1)}\right)=\sum\limits_{j=1}^\infty\left(\frac{1}{j(j+1)}\right)=1$$
Since one of the iterated series is abosult convergent, cauchys product rule implies that the double series $\sum\limits_{j,k=2}^\infty\frac{1}{j^k}$ is also absolute convergent.
Cauchys product rule also states in that case that:
$$\sum\limits_{j,k=2}^\infty\frac{1}{j^k}=\sum\limits_{j=2}^\infty\left(\sum\limits_{k=2}^\infty\frac{1}{j^k}\right)=1$$

Would be great if someone could look over it and give me feedback if my work is correct , thanks alot :)
 A: 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.
A: There's a nice upper bound: note that
$$
\sum_{j=2}^{\infty}\frac{1}{j^k} <\int_{1}^{\infty}\frac{dx}{x^k}
$$
by continuity of $\frac{1}{x^k}$ in the vicinity of $1, \exists \varepsilon$ s.t.
$$
\sum_{j=2}^{\infty}\frac{1}{j^k} < \int_{1 + \varepsilon}^{\infty}\frac{dx}{x^k} = \frac{{t}^{k-1}}{k-1}
$$
where $t = \frac{1}{1+\varepsilon}<1$
Then the sum on $k$ converges:
$$
\sum_{k=2}^{\infty}\frac{t^{k-1}}{k-1} = \log (1+\frac{1}{\varepsilon})
$$
