The diagram above comes from a derivation of the formula for the volume of a cone; it's one of the preliminary steps, and sadly, I'm stuck on it.
What we're doing here is inscribing an "infinite" number of equally thick circular discs in the cone to arrive at the formula for the cone's volume.
We are to let $n$ be a given large positive integer. The text then says: "if there are $n-1$ inscribed discs, then each disc has thickness $h \over n$, as shown in the figure."
It seems to me, however, that if we have a total of $n-1$ discs, the thickness of each disc is $h \over n-1$ and not $h \over n$. I confirmed this for myself looking at a couple of simple examples (admittedly with $n$ failing to be a "large" integer, but surely the truth of the claim made in the diagram shouldn't be affected by that).
What am I missing? Thanks in advance for any/all guidance here.
I suspect the discrepancy is about the top disk in the pile - in particular, suppose you wanted to place disks of thickness $\frac{h}3$ in the cone. Since the text asks for the disks to be inscribed, we find that the disk resting on the base of the cone has radius $\frac{2}3r$, since it extends up to a height of $\frac{h}3$ above the base, where the cone has that radius. The disk atop that would have a radius of $\frac{1}3r$. Then, the disk atop that would have a radius of $0$, since it reaches to the top of the cone, where the radius goes to $0$.
So, you could either could $2$ disks, excluding the top one, or count $3$ disks, but have one of them be degenerate. Presumably, the next step is to sum up the volumes of these disks, in which case it doesn't matter whether you included or excluded the top disk of volume zero.