Summation by parts (Tag PR-2MTPE)
Let \((a_k)_{k\ge1}\) and \((b_k)_{k\ge1}\) be sequences of complex numbers, and put \(A_n \coloneqq \sum_{k=1}^n a_k\) and \(B_n\coloneqq\sum_{k=1}^n b_k\) for \(n\ge1\), with \(A_0 \coloneqq 0\) and \(B_0\coloneqq0\).
For all integers \(1\le m\le n\), \[ \sum_{k=m}^n a_k b_k = A_nb_n - A_{m-1} b_m - \sum_{k=m}^{n-1} A_k(b_{k+1} - b_{k}). \]
For all sequences \((f_k)\) and \((g_k)\) of complex numbers and all integers \(m\le n\), writing \(\Delta g_k \coloneqq g_{k+1} - g_k\) and \(\Delta f_k\coloneqq f_{k+1}-f_k\), \[ \sum_{k=m}^n f_k \, \Delta g_k = f_{n+1} g_{n+1} - f_m g_m - \sum_{k=m}^n g_{k+1} \, \Delta f_k. \]
For every \(n\ge1\), \[ A_{n} B_{n}=\sum_{k=1}^{n} A_{k} b_{k}+\sum_{k=1}^{n} a_{k} B_{k-1}. \]
For (b), \(f_{k+1}g_{k+1}-f_kg_k=f_k\,\Delta g_k+g_{k+1}\,\Delta f_k\) for every \(k\); summing over \(m\le k\le n\) telescopes the left side to \(f_{n+1}g_{n+1}-f_mg_m\).
For (a), substitute \(a_k=A_k-A_{k-1}\): \[ \sum_{k=m}^n a_kb_k=\sum_{k=m}^nA_kb_k-\sum_{k=m-1}^{n-1}A_kb_{k+1}=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k(b_{k+1}-b_k). \]
For (c), \(A_kB_k-A_{k-1}B_{k-1}=A_k(B_{k-1}+b_k)-(A_k-a_k)B_{k-1}=A_kb_k+a_kB_{k-1}\) for every \(k\ge1\); summing over \(1\le k\le n\) and using \(A_0B_0=0\) gives the identity.
Identity (a) is the discrete analogue of integration by parts: for \(F\) an antiderivative of \(f\) on \([a,b]\) and \(g\) continuously differentiable, \[ \int_a^b f g = F(b)g(b) - F(a)g(a) - \int_a^b Fg'. \] The partial sums \(A_k\) play the role of \(F\), and the differences \(b_{k+1}-b_k\) the role of \(g'\).
Let \(A_n \coloneqq \sum_{k\leq n} a_k\), so that \(a_k = A_k - A_{k-1}\). Reindexing the second sum and separating its boundary terms, \[ \begin{aligned} \sum_{m\leq k \leq n} a_k b_k &= \sum_{m\leq k \leq n} (A_k - A_{k-1}) b_k \\ &= \sum_{m\leq k \leq n} A_kb_k - \sum_{m-1\leq k \leq n-1} A_{k} b_{k+1} \\ &= A_nb_n - A_{m-1} b_{m} + \sum_{m\leq k \leq n-1} A_kb_k - \sum_{m\leq k \leq n-1} A_{k} b_{k+1} \\ &= A_nb_n - A_{m-1} b_{m} - \sum_{m\leq k \leq n-1} A_k(b_{k+1} - b_{k}). \end{aligned} \]
Dirichlet's test (Tag L-EAZX6)
Let \((a_k)_{k\ge1}\) and \((b_k)_{k\ge1}\) be sequences of real numbers, and let \(A_n\coloneqq\sum_{k=1}^n a_k\). Suppose that
- the sequence of partial sums \((A_n)_{n\ge1}\) is bounded, and
- \((b_k)\) is nonincreasing and \(b_k\to0\).
Then the series \(\sum_{k\geq 1} a_k b_k\) converges.
Let \(\abs{A_n}\leq M\) for all \(n\), where \(A_n \coloneqq \sum_{k\leq n} a_k\) and \(A_0 \coloneqq 0\). By summation by parts with \(m=1\), \[ \sum_{k=1}^N a_k b_k = A_N b_N + \sum_{k=1}^{N-1} A_k (b_k - b_{k+1}). \] Since \(\abs{A_N b_N}\leq M b_N \to 0\), the first term tends to \(0\). Since \(b_k\) is decreasing, \(b_k - b_{k+1}\geq 0\), and \[ \sum_{k=1}^{N-1}\abs{A_k (b_k - b_{k+1})} \leq M\sum_{k=1}^{N-1} (b_k - b_{k+1}) = M(b_1 - b_N) \leq M b_1, \] so the second sum converges absolutely as \(N\to\infty\). Hence \(\sum_{k\geq 1} a_k b_k\) converges.
Abel's theorem (Tag T-B7YTE)
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 \(r\in(0,1)\).
Abel's test for power series on the unit circle (Tag L-MYZOX)
Let \((c_k)_{k\ge0}\) be a nonincreasing sequence of nonnegative real numbers with \(c_k\to0\), and let \(f(z)\coloneqq\sum_{k\ge0}c_kz^k\). Then the series \(f(z)\) converges for every \(z\) with \(\abs{z}=1\) and \(z\neq1\).
Let \(\abs{z}=1\) with \(z\neq1\). The partial sums of \(\sum_k z^k\) satisfy \(\abs{\sum_{k=0}^{N}z^k}=\abs{\frac{1-z^{N+1}}{1-z}}\le\frac{2}{\abs{1-z}}\) for every \(N\), so they are bounded. Since \((c_k)\) is nonincreasing with limit \(0\), Dirichlet's test applied to \(a_k=c_k\) and \(b_k=z^k\) shows that \(\sum_k c_kz^k\) converges.
(The alternating harmonic series)
Integrating the geometric series \(\sum_{k\geq 0}(-z)^k = 1/(1+z)\) term by term gives \[ \sum_{k\geq 1} {(-1)^{k+1} z^k \over k} = \log(1+z), \qquad \abs{z} < 1. \] The series converges at \(z=1\) by the alternating series test, so Abel's theorem gives \(\sum_{k\geq 1}(-1)^{k+1}/k = \lim_{x\to 1^-}\log(1+x) = \log 2\).
(The converse of Abel's theorem fails)
Let \(f(z) \coloneqq \sum_{n\geq 0} (-z)^n = 1/(1+z)\) for \(\abs z<1\). Then \(\lim_{x\to 1^-} f(x) = 1/2\), but the series \(\sum_{n\geq 0}(-1)^n = 1-1+1-\cdots\) diverges.
Exercises
Convergence of $\sum kz^k$, $\sum z^k/k^2$, and $\sum z^k/k$ on the unit circle (Tag E-SS1.EX-19)
Show that
\(\sum kz^k\) diverges on \(S^1\).
\(\sum k^{-2} z^k\) converges on \(S^1\).
\(\sum \inverseof{k} z^k\) converges on \(S^1\sm\ts{1}\) and diverges at \(1\).
Solution
<1>1. (1) \(\sum kz^k\) diverges at every \(z\in S^1\).
The terms of a convergent series tend to \(0\), but \(\abs{kz^k} = k \to \infty\) for \(\abs z=1\).
<1>2. (2) \(\sum k^{-2} z^k\) converges at every \(z\in S^1\).
For \(\abs z=1\), \(\sum \abs{k^{-2} z^k} = \sum k^{-2}\) converges by the \(p\)-test, and an absolutely convergent series converges.
<1>3. (3) \(\sum k^{-1} z^k\) diverges at \(z=1\) and converges at every \(z\in S^1\setminus\{1\}\).
At \(z=1\) the series is the harmonic series. Otherwise \(z = e^{i\theta}\) with \(\theta \in (0, 2\pi)\). The partial sums of \(\sum e^{ik\theta}\) are bounded, since the geometric sum formula gives \[\abs{\sum_{k=0}^{m} e^{ik\theta}} = \abs{\frac{1 - e^{i(m+1)\theta}}{1 - e^{i\theta}}} \leq \frac{2}{\abs{1- e^{i\theta}}}\] for every \(m\), using \(\abs{1 - e^{i(m+1)\theta}}\le2\). The coefficients \(1/k\) decrease monotonically to \(0\), so Dirichlet’s test gives convergence.
Convergence of $\prod_{n\in\mathbb{Z}}(1+a_n)$ when $\{a_n\}\in\ell_1(\mathbb{Z})$ (Tag E-ORJPT)
Show that \(\prod_{n\in \ZZ} (1 + a_n) < \infty\) if \(\ts{a_n} \in \ell_1(\ZZ)\).
Solution
<1>1. Since \(\{a_n\} \in \ell_1(\ZZ)\), we have \(\sum_{n \in \ZZ} |a_n| < \infty\).
definition of \(\ell_1\).
<1>2. Hence \(a_n \to 0\), so for all sufficiently large \(|n|\) we have \(|a_n| < 1/2\).
a convergent series has terms tending to \(0\).
<1>3. For \(|a_n| < 1/2\), \(|\log(1 + a_n)| \le 2|a_n|\).
\(|\log(1+z)| \le 2|z|\) for \(|z| \le 1/2\) (standard estimate).
<1>4. Hence \(\sum_{n} |\log(1 + a_n)| \le 2\sum_n |a_n| < \infty\).
<1>3 and <1>1.
<1>5. Therefore \(\sum_n \log(1 + a_n)\) converges absolutely, so the product \(\prod_n (1 + a_n)\) converges to a finite nonzero value.
a product \(\prod (1 + a_n)\) converges (absolutely) iff \(\sum \log(1 + a_n)\) converges (absolutely).
<1>6. Q.E.D.
<1>5.
Application of summation by parts (Tag E-XIT36)
Use summation by parts to show that \(\sin(n)/n\) converges.
Solution
As written, \(\sin(n)/n\) is a sequence rather than a series, and its convergence is immediate: \[ \left|{\sin n\over n}\right|\le {1\over n}\longrightarrow0. \] Thus the displayed sequence converges to \(0\); summation by parts is not needed for this assertion.
The standard summation-by-parts exercise suggested by the wording is the convergence of \[ \sum_{n=1}^\infty {\sin n\over n}. \] For that corrected statement, put \[ B_N=\sum_{n=1}^N\sin n. \] Using a geometric sum, \[ \sum_{n=1}^N e^{in} =e^i{1-e^{iN}\over1-e^i}, \] so \((B_N)\) is bounded. Summation by parts gives, for \(N>M\), \[ \sum_{n=M}^N {\sin n\over n} ={B_N\over N}-{B_{M-1}\over M} +\sum_{n=M}^{N-1}B_n\left({1\over n}-{1\over n+1}\right). \] If \(|B_n|\le C\), the absolute value of the right-hand side is at most \[ {C\over N}+{C\over M} +C\sum_{n=M}^{N-1}\left({1\over n}-{1\over n+1}\right) \le {3C\over M}, \] which tends to \(0\) as \(M\to\infty\). Hence the corrected series is Cauchy and therefore convergent.
Convergence of $\frac1z\sum_{k=1}^\infty\frac{z^k}{k}$ on $S^1\setminus\{1\}$ by summation by parts (Tag E-FJZCL)
Show that \(\frac{1}{z}\sum_{k=1}^\infty \frac{z^k}{k}\) converges for all \(z \in S^1 \setminus \{1\}\) using summation by parts.
Solution
Fix \(z\in S^1\setminus\{1\}\) and set \[ A_N=\sum_{k=1}^N z^k. \] Since \[ A_N=z\frac{1-z^N}{1-z}, \] we have the uniform-in-\(N\) bound \[ |A_N|\le \frac{2}{|1-z|}. \]
Summation by parts gives \[ \sum_{k=1}^N\frac{z^k}{k} =\frac{A_N}{N} +\sum_{k=1}^{N-1}A_k\left(\frac1k-\frac1{k+1}\right). \] The first term tends to \(0\), while \[ \sum_{k=1}^\infty |A_k|\left(\frac1k-\frac1{k+1}\right) \le \frac{2}{|1-z|} \sum_{k=1}^\infty\frac1{k(k+1)}<\infty. \] Thus \(\sum z^k/k\) converges. Since \(z\ne0\), multiplication by \(1/z\) preserves convergence.