In 3-dimensional chess, it is possible to force checkmate starting with a finite number of rooks. As this fact still appears to be open, I'll post a method of forcing checkmate with 96 rooks, even though it should be clear that this is not optimal. You can remove some of the rooks from the method I'll give below, but I am aiming for a simple explanation of the method rather than the fewest possible number of rooks.
First, we move all of the rooks far away in the $z$ direction, so that they cannot be threatened by the king. We also move each of the rooks so that they all have distinct $z$ coordinates. That way, they are free to move any number of steps in the $x$ and $y$ directions without blocking each other. The king will be in check whenever it has the same $(x,y)$-coordinate as one of the rooks. We can project onto the $(x,y)$-plane to reduce it to a 2-dimensional board. Looked at this way, each rook can move any number of places in the $x$ or $y$ direction (and can pass through each other, can pass through the king, and you can multiple rooks in the same $(x,y)$-square). The king is in check if it is on the same square as a rook.
First, I'll describing the following "blocking move" to stop the king passing a given horizontal (or vertical) line.
In the position above, the right-most 3 rooks are stopping the king moving past the red line on the next move. Then, once the king moves, do the following. (i) If the king's $x$-coordinate does not change, do nothing. (ii) If the king's $x$-coordinate increases by one, move the left-most rook so that it is to the right of the other three. Then you are back in the same position, just moved along by one step. (iii) If the king's $x$-coordinate decreases by one step, do nothing. We are back in the same situation, except reflected (so, keep performing the same steps, but reflected in the $x$-direction on subsequent moves).
This way, we chase the king along the red line, but he can never cross it. Furthermore, if the king changes from going right to going left, we have a free move to do something elsewhere on the board. Actually, for this to work, if the king is in column $i$, we just need three rooks at positions $i-1,i,i+1$ on the row above the red line, and one more at any other position on the row. Next, if we have 4 rooks stationed somewhere on the given horizontal row, how many moves does it take to move them into the blocking position? The answer is 6. You first move one rook to have the same $x$-coordinate as the king (say, $x=i$). After the king moves, by reflection we can assume he keeps the same $x$-coordinate or moves one to the right. Then, move the next rook to position $i+2$. Then, after the next move, move a rook to position $i-2$ or $i+4$ in such a way that we have three rooks on the row, with one space between each of them, and the kings is in one of the 3 middle columns. Say, the rooks are at positions $j-2,j,j+2$ and the king is in column $j-1,j$ or $j+1$. If the king moves to column $j-1$ or $j+1$ we just move the 4th rook to this position and we have attained the blocking position. If the king moves to column $j$, we move the rook in position $j-2$ to position $j-1$ and, on the next move, we can move the 4th rook in to attain the blocking position. If the king moves to column $j+2$, we move the rook in column $j-2$ to $j+4$, then we are in the position above where there are rooks at positions $k-2,k,k+2$ and the king in position $k$, so it takes 2 more moves to attain the blocking position.
So, we just need to keep 4 rooks stationed along the row which we wish to block the king from crossing. Whenever he moves within 6 steps from this row, start moving the rooks into the blocking position, and he can never step into the given row.
Now, choose a large rectangle surrounding the king, and position 15 rooks in each corner as below.
Also, position 4 rooks in arbitrary positions along each edge of the rectangle. So, that's $4\times15+4\times4=76$ rooks used so far. I puposefully left some of the board blank in the diagram above. The point is to not specify exactly how big the rectangle is. It doesn't matter, just so long as it is large enough to be able to move the 76 rooks into position before the black king can get within 6 steps of any of the edges of the rectangle.
Now, once we are in this position, then whenever the black king moves within one of the red rectangles, use the 4 rooks positioned along the adjacent edge to perform the blocking move as described above to stop the king crossing that edge. We can keep doing this, and imprison the black king within the big rectangle. Furthermore, we keep getting free moves to do something else whenever the king moves out of the red rectangles, or whenever he changes direction within a red rectangle. Also, if the king is in one of the inside corners of a red rectangle, there is already a rook in the corresponding position at the adjacent edge of the big rectangle, giving us a free move.
Now, suppose that we have an extra 20 rooks. During the free moves we get while chasing the king around the edge of the big square, we can move these to any position we like. With 20 rooks, we can position 16 of to the left of each of the 16 rooks near the right-hand corners of the big rectangle which have an empty square to their left. Also, position 4 rooks along the column one step to the left of the right-hand edge of the big rectangle. This way, we create a new rectangle one square smaller in the $x$-direction. If the king ever enters the right-hand red rectangle or one step to the left of this, we use the new 4 blocking rooks to stop him from reaching the right-hand edge of the new big rectangle. If he is already within the red rectangle, and stays there then, when we get a free move, we can move one of the new blocking rooks to the position one above or below the row in which the king is. Then we can bring the other 3 rooks in, blocking him out of this column. In this way, we create a new big rectangle one step smaller in the $x$-direction and with the king still trapped inside. Similarly, we can reduce the height of the big rectangle by 1. Repeat this, enclosing the king in ever smaller rectangles until, eventually, he gets trapped in the single square within a 3x3 rectangle surrounded by 8 rooks. Then bring one of the other rooks in to cover this square, which is checkmate.
Best Answer
Queens. Two queens suffice, as pointed out in the comments of Sp3000.
Rooks. Three rooks suffice, by trapping the black king between two walls, and then using the third rook to slowly close the gap between the walls. If the black king approach a rook, simply move it on the same file but further away. Two rooks do not suffice, since there is no check-mate position with two white rooks and a black king.
Bishops. Six bishops suffice, with three of each color. White can use a pair of bishops to form a wall, and thus with two pairs white can trap the black king between two walls, which gradually close and deliver checkmate with the fifth or sixth (whichever color is needed). If the black king approaches a bishop, simply move it away on the same diagonal. Five bishops do not suffice, since the black king can simply stay on the color having at most two bishops of that color, and there will always be a square available, since double check will not arise.
Knights. No finite number of white knights will suffice in this game. To see this, observe first that two white knights do not suffice, since one cannot set up a check-mated position. Now, suppose white places finitely many knights on the board. Black need only place his king on a file completely to the right of those pieces. Black's strategy is simply to move steadily to the right (that is, his every move should involve a rightward horizontal movement). Since knights move at most two units to the right, white will not be able to keep more than two knights in the vicinity of the black king, and indeed, will not be able to bring two knights so close. With one knight, to be sure, white can move at double speed and catch up to the black king; but to bring two knights close to the king, each would have had to move at a greater than one unit to the right speed, which is not possible. Finally, notice that with infinitely many knights, we can fill up the board almost completely, leaving only a few empty spots, and easily set a trap for black this way.
Conclusion. So we have $Q=2$, $R=3$, $B=6$ and $N=\infty$, giving $$\frac{1}{Q}+\frac1R+\frac1B+\frac1N=\frac12+\frac13+\frac16+\frac1{\infty}=1.$$