If $\sum c_k$ converges, then $\sum c_kr^k\to\sum c_k$ as $r\to1^-$.Suggest a slogan
Let \((c_k)_{k\ge0}\) be complex numbers such that \(\sum_{k\ge0}c_k\) converges. Then \(\sum_{k\ge0}c_kz^k\) converges for \(\abs z<1\), and \[ \lim_{r\to1^-}\sum_{k\ge0}c_kr^k=\sum_{k\ge0}c_k, \] where the limit is taken over real \(r\in(0,1)\).