Given two sequences of real numbers \(\left\{{ a_k }\right\} , \left\{{ b_k }\right\}\) which satisfy
- The sequence of partial sums \(\left\{{ A_n }\right\}\) is bounded,
- \(b_k \searrow 0\).
then \begin{align*} \sum_{k\geq 1} a_k b_k < \infty .\end{align*}
Use summation by parts. For a fixed \(\sum a_k b_k\), write \begin{align*} \sum_{n=1}^m x_n Y_n + \sum_{n=1}^m X_n y_{n+1} = X_m Y_{m+1} .\end{align*} Set \(x_n \coloneqq a_n, y_N \coloneqq b_n - b_{n-1}\), so \(X_n = A_n\) and \(Y_n = b_n\) as a telescoping sum. Importantly, all \(y_n\) are negative, so \({\left\lvert {y_n} \right\rvert} = {\left\lvert {b_n - b_{n-1}} \right\rvert} = b_{n-1} - b_n\), and moreover \(a_n b_n = x_n Y_n\) for all \(n\). We have \begin{align*} \sum_{n\geq 1} a_n b_n &= \lim_{N\to\infty} \sum_{n\leq N} x_n Y_n \\ &= \lim_{N\to\infty} \sum_{n\leq N} X_N Y_N - \sum_{n\leq N} X_n y_{n+1} \\ &= - \sum_{n\geq 1} X_n y_{n+1}, \end{align*} where in the last step we’ve used that \begin{align*} {\left\lvert {X_N} \right\rvert} = {\left\lvert {A_N} \right\rvert}\leq M \implies {\left\lvert {X_N Y_{N} } \right\rvert} = {\left\lvert {X_N} \right\rvert} {\left\lvert {b_{n+1}} \right\rvert} \leq M b_{n+1} \to 0 .\end{align*} So it suffices to bound the latter sum: \begin{align*} \sum_{k\geq n}{\left\lvert { X_k y_{k+1} } \right\rvert} &\leq M \sum_{k\geq 1} {\left\lvert {y_{k+1}} \right\rvert}\\ &\leq M \sum_{k\geq 1} b_{k} - b_{k+1} \\ &\leq 2M(b_1 - b_{n+1})\\ &\leq 2M b_1 .\end{align*}
If \(\sum_{k=1}^\infty c_k z^j\) converges on \({\left\lvert {z} \right\rvert} < 1\) then \begin{align*} \lim_{z\to 1^-} \sum_{k\in {\mathbb{N}}} c_k z^k = \sum_{k\in {\mathbb{N}}} c_k .\end{align*}
If \(f(z) \coloneqq\sum c_k z^k\) is a power series with \(c_k \in {\mathbf{R}}^{\geq 0}\) and \(c_k\searrow 0\), then \(f\) converges on \(S^1\) except possibly at \(z=1\).
What is the value of the alternating harmonic series? Integrate a geometric series to obtain \begin{align*} \sum {(-1)^k z^k \over n} = \log(z+1) && {\left\lvert {z} \right\rvert} < 1 .\end{align*} Since \(c_k \coloneqq(-1)^k/k \searrow 0\), this converges at \(z=1\), and by Abel’s theorem \(f(1) = \log(2)\).
The converse to Abel’s theorem is false: take \(f(z) = \sum (-z)^n = 1/(1+z)\). Then \(f(1) = 1-1+1-\cdots\) diverges at 1, but \(1/1+1 = 1/2\). So the limit \(s\coloneqq\lim_{x\to 1^-} f(x) 1/2\), but \(\sum a_n\) doesn’t converge to \(s\).
Setting \(A_n \coloneqq\sum_{k=1}^n b_k\) and \(B_0 \coloneqq 0\), \begin{align*} \sum_{k=m}^n a_k b_k &= A_nb_n - A_{m-1} b_m - \sum_{k=m}^{n-1} A_k(b_{k+1} - b_{k}) .\end{align*} Compare this to integrating by parts: \begin{align*} \int_a^b f g = F(b)g(b) - F(a)g(a) - \int_a^b Fg' .\end{align*} How to remember: set \(\Delta g_k \coloneqq g_{k+1} - g_k\) and \(\mathbf{I} g_k = g_{k+1}\), then \begin{align*} \sum_{k=m}^n f_k \cdot \Delta g_k = f_{n+1} g_{n+1} - f_m g_m - \sum_{k=m}^n \mathbf{I}g_{k} \cdot \Delta f_k .\end{align*}
Note there is a useful form for taking the product of sums: \begin{align*} A_{n} B_{n}=\sum_{k=1}^{n} A_{k} b_{k}+\sum_{k=1}^{n} a_{k} B_{k-1} .\end{align*}
An inelegant proof: define \(A_n \coloneqq\sum_{k\leq n} a_k\), use that \(a_k = A_k - A_{k-1}\), reindex, and peel a top/bottom term off of each sum to pattern-match.\
Behold: \begin{align*} \sum_{m\leq k \leq n} a_k b_k &= \sum_{m\leq k \leq n} (A_k - A_{k-1}) b_k \\ &= \sum_{m\leq k \leq n} A_kb_k - \sum_{m\leq k \leq n} A_{k-1} b_k \\ &= \sum_{m\leq k \leq n} A_kb_k - \sum_{m-1\leq k \leq n-1} A_{k} b_{k+1} \\ &= A_nb_n + \sum_{m\leq k \leq n-1} A_kb_k - \sum_{m-1\leq k \leq n-1} A_{k} b_{k+1} \\ &= A_nb_n - A_{m-1} b_{m} + \sum_{m\leq k \leq n-1} A_kb_k - \sum_{m\leq k \leq n-1} A_{k} b_{k+1} \\ &= A_nb_n - A_{m-1} b_{m} + \sum_{m\leq k \leq n-1} A_k(b_k - b_{k+1}) \\ &= A_nb_n - A_{m-1} b_{m} - \sum_{m\leq k \leq n-1} A_k(b_{k+1} - b_{k}) .\end{align*}
Exercises
Show that
- \(\sum kz^k\) diverges on \(S^1\).
- \(\sum k^{-2} z^k\) converges on \(S^1\).
- \(\sum k^{-1}z^k\) converges on \(S^1\setminus\left\{{1}\right\}\) and diverges at \(1\).
- Use that \({\left\lvert {z^k} \right\rvert} = 1\) and \(\sum c_kz^k < \infty \implies {\left\lvert {c_k} \right\rvert} \to 0\), but \({\left\lvert {kz^k} \right\rvert} = {\left\lvert {k} \right\rvert} \to \infty\) on \(S^1\).
- Use that absolutely convergent implies convergent, and \(\sum {\left\lvert {k^{-2} z^k} \right\rvert} = \sum {\left\lvert {k^{-2}} \right\rvert}\) converges by the \(p{\hbox{-}}\)test.
- If \(z=1\), this is the harmonic series. Otherwise take \(a_k = 1/k, b_k = e^{i k \theta}\) where \(\theta \in (0, 2\pi)\) is some constant, and apply Dirichlet’s test. It suffices to bound the partial sums of the \(b_k\). Recalling that \(\sum_{k\leq N} r^k = (1-r^{N+1}) / (1-r)\), \begin{align*} {\left\lVert { \sum_{k\leq m} e^{ik\theta } } \right\rVert} = {\left\lVert {1 - e^{i(m+1)\theta} \over 1 - e^{i\theta}} \right\rVert} \leq {2 \over {\left\lVert { 1- e^{i\theta}} \right\rVert}} \coloneqq M ,\end{align*} which is a constant. Here we’ve used that two points on \(S^1\) are at most distance 2 from each other.
Show that \(\prod_{n\in {\mathbf{Z}}} (1 + a_n) < \infty\) if \(\left\{{a_n}\right\} \in \ell_1({\mathbf{Z}})\).
Use summation by parts to show that \(\sin(n)/n\) converges.
Show that \({1\over z} \sum_{k=1}^\infty {z^k \over k}\) converges on \(S^1 \setminus\left\{{1}\right\}\) using summation by parts.