Suppose \(\sum_{n=1}^{\infty}a_n\) converges. Prove that \[ \lim_{r\to1^-}\sum_{n=1}^{\infty}r^na_n = \sum_{n=1}^{\infty}a_n. \] In other words, a convergent series is Abel summable to its ordinary sum.
Solution
1. Let\[ s_n=\sum_{k=1}^n a_k, \qquad s=\sum_{k=1}^{\infty}a_k. \]Then \((s_n)\) is bounded and \(s_n\to s\).
Convergence of the series means exactly that its sequence of partial sums converges to \(s\). Every convergent sequence is bounded.
2. For \(0\le r<1\) and every \(N\ge1\),\[ \sum_{n=1}^N a_nr^n =s_Nr^N+(1-r)\sum_{n=1}^{N-1}s_nr^n. \]
Since \(a_n=s_n-s_{n-1}\) with \(s_0=0\), \[ \begin{aligned} \sum_{n=1}^N a_nr^n &=\sum_{n=1}^N(s_n-s_{n-1})r^n\\ &=\sum_{n=1}^Ns_nr^n-r\sum_{n=0}^{N-1}s_nr^n\\ &=s_Nr^N+(1-r)\sum_{n=1}^{N-1}s_nr^n. \end{aligned} \] This is the summation-by-parts formula of E-SS1.EX-14 with the sequences \(r^n\) and \(a_n\) in the roles of \(a_n\) and \(b_n\).
3. For every \(0\le r<1\),\[ \sum_{n=1}^{\infty}a_nr^n =(1-r)\sum_{n=1}^{\infty}s_nr^n. \]
By 1 there is \(C\) with \(|s_n|\le C\). Hence \(|s_Nr^N|\le Cr^N\to0\) and the series \(\sum s_nr^n\) converges absolutely. Letting \(N\to\infty\) in 2 gives the identity.
4. For \(0\le r<1\),\[ (1-r)\sum_{n=1}^{\infty}s_nr^n =sr+(1-r)\sum_{n=1}^{\infty}(s_n-s)r^n. \]
Write \(s_n=s+(s_n-s)\) and use \[ (1-r)\sum_{n=1}^{\infty}r^n=r. \]
5. One has\[ \lim_{r\to1^-}(1-r)\sum_{n=1}^{\infty}(s_n-s)r^n=0. \]
Fix \(\varepsilon>0\). Choose \(N\) so that \(|s_n-s|<\varepsilon\) for every \(n\ge N\). Then \[ \begin{aligned} (1-r)\sum_{n=1}^{\infty}|s_n-s|r^n &\le (1-r)\sum_{n=1}^{N-1}|s_n-s|r^n +\varepsilon(1-r)\sum_{n=N}^{\infty}r^n\\ &\le (1-r)\sum_{n=1}^{N-1}|s_n-s| +\varepsilon. \end{aligned} \] The first term tends to \(0\) as \(r\to1^-\) because it is \((1-r)\) times a fixed finite constant. Thus the limsup is at most \(\varepsilon\). Since \(\varepsilon\) is arbitrary, the asserted limit is \(0\).
6. Therefore\[ \lim_{r\to1^-}\sum_{n=1}^{\infty}a_nr^n=s. \]
By 3 and 4, \[ \sum_{n=1}^{\infty}a_nr^n =sr+(1-r)\sum_{n=1}^{\infty}(s_n-s)r^n. \] The first term tends to \(s\), and the second tends to \(0\) by 5.