$m_i>2k$ for all $i=0,\ldots,u-1$. Analogously to the proof of Proposition \[drad\_2k\_prop\], we get $\left\lfloor\frac{3m_i}{4}\right\rfloor\leq m_{i+1}$ and using Lemma \[rad\_2k\_phaselem\], we deduce $\operatorname{cost_{rad}}({\mathcal{C}}_{m_{i+1}})<\left(1+6\sqrt[d]{\frac{2k}{m_i}}\right)\cdot\operatorname{cost_{rad}}({\mathcal{C}}_{m_i})+6\sqrt[d]{\frac{2k}{m_i}}\cdot\operatorname{opt}_k$ for all $i=0,\ldots,u-1$. By repeatedly applying this inequality and using $\operatorname{cost_{rad}}({\mathcal{C}}_{2k})\leq\operatorname{cost_{rad}}({\mathcal{C}}_{m_u})$ and $\operatorname{cost_{rad}}({\mathcal{C}}_{m_0})=0$, we get $$\begin{aligned}
\nonumber
\operatorname{cost_{rad}}\left({\mathcal{C}}_{2k}\right)&<\sum_{i=0}^{u-1}\left(6\sqrt[d]{\frac{2k}{m_i}}\cdot\prod_{j=i+1}^{u-1}\left(1+6\sqrt[d]{\frac{2k}{m_j}}\right)\right)\cdot\operatorname{opt}_k\\
\label{rad_2k_prop_mi}
&\leq6\sqrt[d]{\frac{2k}{|X|}}\cdot\sum_{i=0}^{u-1}\left(\left(\frac{4}{3}\right)^\frac{i}{d}\cdot\prod_{j=i+1}^{u-1}\left(1+6\sqrt[d]{\frac{2k}{|X|}}\cdot\left(\frac{4}{3}\right)^\frac{j}{d}\right)\right)\cdot\operatorname{opt}_k\\
\label{rad_2k_prop_renum}
&=6\sqrt[d]{\frac{2k}{|X|}}\left(\frac{4}{3}\right)^\frac{u-1}{d}\cdot\sum_{i=0}^{u-1}\left(\left(\frac{3}{4}\right)^\frac{i}{d}\cdot\prod_{j=u-i}^{u-1}\left(1+6\sqrt[d]{\frac{2k}{|X|}}\cdot\left(\frac{4}{3}\right)^\frac{u-1}{d}\left(\frac{3}{4}\right)^\frac{u-1-j}{d}\right)\right)\cdot\operatorname{opt}_k.\end{aligned}$$ Here, we obtain (\[rad\_2k\_prop\_mi\]) using $m_i\geq\left(\frac{3}{4}\right)^i|X|$ and we obtain (\[rad\_2k\_prop\_renum\]) by substituting $u-1-i$ for $i$. Using $u-1<\log_\frac{4}{3}\frac{|X|}{2k}$, we deduce $$\label{rad_2k_propeq}
\operatorname{cost_{rad}}\left({\mathcal{C}}_{2k}\right)<6\sum_{i=0}^{u-1}\left(\left(\frac{3}{4}\right)^\frac{i}{d}\cdot\prod_{j=0}^{i-1}\left(1+6\left(\frac{3}{4}\right)^\frac{j}{d}\right)\right)\cdot\operatorname{opt}_k.$$ By taking only the first two terms of the series expansion of the exponential function, we get $1+6\left(\frac{3}{4}\right)^\frac{j}{d}<e^{6\left(\frac{3}{4}\right)^\frac{j}{d}}$ and therefore $$\label{rad_2k_prodeq}
\prod_{j=0}^{i-1}\left(1+6\left(\frac{3}{4}\right)^\frac{j}{d}\right)<\prod_{j=0}^{i-1}\left(e^{6\left(\frac{3}{4}\right)^\frac{j}{d}}\right)=e^{6\sum_{j=0}^{i-1}\left(\frac{3}{4}\right)^\frac{j}{d}}.$$ The sum in the exponent can be bounded by the infinite geometric series $$\label{rad_2k_sumeq}
\sum_{j=0}^\infty\left(\frac{3}{4}\right)^\frac{j}{d}<\frac{1}{1-\left(\frac{3}{4}\right)^\frac{1}{d}}\leq4d,$$ where the last inequality follows by upper bounding the convex function $f(x)=\left(\frac{3}{4}\right)^x$ in the interval $[0,1]$ by the line through $f(0)$ and $f(1)$. Putting Inequalities (\[rad\_2k\_propeq\]), (\[rad\_2k\_prodeq\]) and (\[rad\_2k\_sumeq\]) together then gives $$\operatorname{cost_{rad}}\left({\mathcal{C}}_{2k}\right)<6\sum_{i=0}^{u-1}\left(\left(\frac{3}{4}\right)^\frac{i}{d}\cdot e^{24d}\right)\cdot\operatorname{opt}_k<24d\cdot e^{24d}\cdot\operatorname{opt}_k,$$ where the last inequality follows by using Inequality (\[rad\_2k\_sumeq\]) again.

### Connected instances {#connected_sec}

The analysis of the remaining merge steps from the discrete $k$-center case (cf. Section \[drad\_rem\_sec\]) is not transferable to the $k$-center case. Again, as in the proof of Proposition \[rad\_2k\_prop\], we are no longer able to derive a simple additive bound on the increase of the cost when merging 