Convergence implies Abel summability, and we only need to consider when $s=0$? Suppose $\displaystyle c_n\in\mathbb{C}\textrm{ and}\sum_{n=1}^{\infty}c_n=s$. Then, prove $\displaystyle\lim_{r\to 1^{-}}\sum_{n=1}^{\infty}r^{n}c_n=s$.
In my text, the author hinted that: we only need to prove the case when $s=0$.
However, I don't get it. Can somebody explain it for me?
 A: For the $s=0$ case, let $s_{n}=\sum_{k=1}^{n}c_{k}$ for $n \ge 1$ and let $s_{0}=0$. Then $s_{n}-s_{n-1}=c_{n}$ for all $n \ge 1$ and $\lim_{n} s_{n}=0$ by assumption. So the following are absolutely convergent for $0 \le r < 1$:
$$
\begin{align}
     \sum_{n=1}^{\infty}r^{n}c_{n}
        & = \sum_{n=1}^{\infty}r^{n}(s_{n}-s_{n-1}) \\
        & = \sum_{n=1}^{\infty}r^{n}s_{n}-\sum_{n=1}^{\infty}r^{n+1}s_{n} \\
        & = (1-r)\sum_{n=1}^{\infty}r^{n}s_{n}.
\end{align}
$$
For $\epsilon > 0$, there exists $N$ such that $|s_{n}| < \epsilon/2$ whenever $n \ge N$, which implies
$$
         \left|(1-r)\sum_{n=N}^{\infty}r^{n}s_{n}\right| \le (1-r)r^{N}\frac{\epsilon}{2}\sum_{n=0}^{\infty}r^{n}=r^{N}\frac{\epsilon}{2} < \frac{\epsilon}{2},\;\;\; n \ge N.
$$
Therefore,
$$
          \left|(1-r)\sum_{n=1}^{\infty}r^{n}s_{n}\right|
             \le \left|(1-r)\sum_{n=1}^{N-1}r^{n}s_{n}\right|+\frac{\epsilon}{2}.
$$
The sum on the right is bounded by $\epsilon/2$ whenever $1-\delta < r < 1$, provided $\delta$ is appropriately chosen. It follows that
$$
              \lim_{r\uparrow 1}\sum_{n=1}^{\infty}r^{n}c_{n} = 0.
$$
