= Solution
Boundedness at the origin allows only $r^n$ for nonconstant interior modes; decay at infinity allows only $r^{-n}$ outside. Write the common surface potential in its <Fourier series> as $\sum_{n\geq1}v_n(\theta)$, where $v_n=A_n\cos n\theta+B_n\sin n\theta$. Continuity then makes the corresponding interior and exterior factors $(r/R)^n$ and $(R/r)^n$.
Their outward radial derivatives at the surface are $nv_n/R$ inside and $-nv_n/R$ outside. Thus the jump is $-2nv_n/R$. Equating it to minus the prescribed surface-charge mode gives
$$
v_n(\theta)=\frac{R}{2n}(a_n\cos n\theta+b_n\sin n\theta).
$$
Therefore the <potential of a charged cylindrical shell with angular Fourier data> is
$$
\boxed{\Phi(r,\theta)=\sum_{n=1}^\infty\frac{R}{2n}\begin{cases}(r/R)^n,&0\leq r\leq R,\\(R/r)^n,&r\geq R\end{cases}(a_n\cos n\theta+b_n\sin n\theta).}
$$
Either branch gives the same value at $R$. The mean angular mode is zero: outside, neither a constant nor a logarithm can survive the condition at infinity; continuity then forces the interior constant to zero. Consequently $\Phi(0)=0$.
The absence of a mean term in the <surface charge density> is essential. A nonzero mean would require a logarithmic exterior potential and would be incompatible with decay at infinity in this two-dimensional reduction. The normalization follows the stated derivative jump, with no additional permittivity factor introduced.
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