I've been doing calculations on my own, just messing around, with infinite series and their respective geometries. I don't actually know enough to do this correctly but I've been led to believe that the next infinite series that I'm working on would be equivalent to the volume of the space of a right triangular pyramid that is shifted or translated (I don't know the proper term) into the $4$th dimension.
So for example. I translated the area of a right triangle into the third dimension by taking 3 right triangles and lining them up with one another to make a right triangular pyramid. The resulting volume of this is $$\frac{1}{6}\times\text{Area of Base}\times \text{Height of Pyramid}$$
So if you took $4$ right triangular pyramids and matched them up to one another four times in the $4$th dimension, what would the resulting volume be?
I would expect it to be after the pattern of the 4th dimensional cube but just a fraction of it like: $$\frac{wxyz}{n}$$
Basically I need to know what fraction of the $4$th dimensional cube would the volume contain if it was entirely constructed from right triangles that are exactly half of the squares area.
In $3$ dimensions, the formula is $1/3$ the area of the base times the height. I'm not sure you got $1/6.$ Perhaps, you mean $1/6$ a rectangular box enclosing the pyramid. That is, $x$ is one side of the triangular base, $y$ is the height of the triangle, $z$ is the height of the pyramid.
In $4$ dimensions, the volume of a pyramid is $1/4$ times the $3-$dimensional volume of the base, or $xyzw/24$ if you've chosen $x,y,z,w$ to represent the sides of a rectangular box enclosing the pyramid in the same way as in $3$ dimensions.