Tetrahedron circumradius in high dimensions

The standard $n$ simplex is the convex hull in $\mathbb{R}^{n+1}$ of the points $$(1,0,0,\dots,0),\\ (0,1,0,\dots,0)\\ \vdots\\ (0,0,\dots,0,1)$$ That is, the set of all points in $\mathbb{R}^{n+1}$ whose coordinates are nonnegative and add up to $1$. The edge length is $\sqrt2$ and the center is at $(\frac1{n+1},\frac1{n+1}, \dots,\frac1{n+1})$. A simple calculation shows that the distance from the center to a vertex approaches $1$ as $n\to\infty,$ so the ratio of the edge length to the circumradius actually goes to $\sqrt2$ as $n\to\infty$.

Thanks to Adam Zalcmann for pointing out the error in my previous calculation. See his comment below for the detailed calculation.

I think the easiest way to go about this problem is to embed the $n-1$ simplex in a plane in $\Bbb{R}^{n}$. If we do this, it is well known that we can represent the vertices of the $n-1$ simplex in $\Bbb{R}^{n}$ as simply $\{\mathbf{e}_i\}$, the unit vectors. However, one might notice that that the mean position of the vertices is $$\frac{1}{n}\sum_{i=1}^n\mathbf{e}_i=\left(\frac{1}{n},...,\frac{1}{n}\right)$$ So it might make sense to shift our coordinates system so that the "center of mass" is the origin. That is, we take the vertices to be $$\mathbf{v}_i=\frac{n-1}{n}\mathbf{e}_i-\frac{1}{n}\sum_{j\in\{1,...,n\}\backslash\{i\}}\mathbf{e}_j$$ However, after doing this one might notice that the side lengths are $$\Vert\mathbf{v}_i-\mathbf{v}_j\Vert=\sqrt{2\left(\frac{n-1}{n}-\frac{-1}{n}\right)^2}=\sqrt{2}$$ So we scale down by this factor and let the vertices of our simplex be $$\mathbf{u}_i=\frac{1}{n\sqrt{2}}\left((n-1)\mathbf{e}_i-\sum_{j\in\{1,...,n\}\backslash\{i\}}\mathbf{e}_j\right)$$ Great. So now the circumradius of the $\boldsymbol{n-1}$ simplex is simply $$R_{n-1}=\Vert\mathbf{u}_i\Vert=\frac{1}{n\sqrt{2}}\sqrt{(n-1)^2+(n-1)}$$ So the circumradius of the $n$ simplex is $$R_n=\frac{\sqrt{n^2+n}}{(n+1)\sqrt{2}}$$ The first few terms in the sequence are $$\{0,1/2,1/\sqrt{3},\sqrt{6}/4,\sqrt{2/5},...\}$$ We see that $R_n\to\frac{1}{\sqrt{2}}\text{ as }n\to\infty$.

The upshot is that the circumradius tends to $(\sqrt{2})^{-1}$ times the edge length.