**Sketching a diagram of the joint distribution might firm up your understanding as well as help you get the right answer (and spot incorrect answers that might be offered).**

Usually you don't need to work very hard at this--it's primarily a conceptual exercise--but for accuracy I asked a computer to draw this diagram:

[![Figure][1]][1]

This diagram situates a dot at each point with coordinates $(n,k)$ where the probability is nonzero.  (The dots are colored and sized in proportion to their probabilities.)  To do that, it creates *vertical strips* of dots, because each such strip corresponds to an event with a single value of $n.$  Those strips therefore reflect the conditional probability information, which says the strip positioned over a whole number $n$ must have dots at heights $k=1,2,\ldots, 2n$ (all with equal probabilities within each strip).  I have highlighted the strip for $n=4$ because that corresponds to the work attempted in the question.

To the right of each dot I have posted a formula for the *joint* (not the conditional) probability.  Recall that the joint probability at $(n,k)$ must be the product of

* The probability of $n,$ given by $1/2\times n \times 2^{-n},$ and

* The conditional probability of $k$ given $n.$  Because this is assumed uniform and it covers $2n$ possibilities, this conditional probability is $1/(2n).$

Thus **the formula for the joint probability is**

$$P(n,k) = P(n)\times P(k\mid n) = \left\{\begin{array}{lr}\frac{1}{2}\,n\,2^{-n}\,/\,(2n) & 1 \le k \le 2n \\ 0 & \text{otherwise.}\end{array}\right.$$

Notice that **the expression on the right hand side simplifies:**

$$\frac{1}{2}\,n\,2^{-n}\,/\,(2n) = 2^{-(n+2)}.$$

You can spot-check these values against the diagram if you wish.

----

**As a reality check,** let's verify the probabilities sum to unity--but *we'll do this in a different way than we constructed the diagram so that the check might catch any mistakes we might have made.*  Let's sum the probabilities by rows:

* At the bottom two rows with $k=1$ and $k=2,$ which are identical, you can read off the sequence of probabilities left to right as $2^{-1-2}, 2^{-2-2}, 2^{-3-2}, \ldots = 1/8, 1/16, 1/32, \ldots.$  This is a geometric series that sums (obviously) to $1/4.$  The two rows together sum to $1/2.$

* At the next two rows with $k=3$ and $k=4,$ which are identical, the sequence of probabilities is the same as before, *but with the first one omitted.*  We obtain two rows summing to $1/16 + 1/32 + 1/64 + \cdots = 1/8.$  The two rows together sum to $1/4.$

* The pattern is evident: every time you go up two rows you see the same probabilities as before but (a) multiplied by $1/2$ and (b) shifted one unit to the right.  Thus the next two sum to $1/4\times 1/2 = 1/8,$ the next two sum to $1/8\times 1/2=1/16,$ and so on.

Evidently the sum of all the probabilities is $1/2 + 1/4 + 1/8 + \cdots = 1,$ as it should be.

-----

**As a mathematical proposition**, this diagram has shown how to evaluate the sum

$$\sum_{n=1}^\infty n\, 2^{-n} = 2$$

by splitting each term $n 2^{-n}$ into $2n$ separate pieces of size $2^{-(n+1)}$ *and then adding those pieces in a different order.*  The evaluation requires knowing only that $1/2+1/4+1/8+\cdots = 1.$


  [1]: https://i.sstatic.net/xMnCa.png