lemma (Dirichlet's Test):

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 $\begin{align*} \sum_{k\geq 1} a_k b_k < \infty .\end{align*}$

proof (?):

See http://www.math.uwaterloo.ca/~krdavids/Comp/Abel.pdf

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: $\begin{align*} \sum_{k\geq n}{\left\lvert { X_k y_{k+1} } \right\rvert} &\leq M \sum_{k\geq 1} {\left\lvert {y_{k+1}} \right\rvert}\\ &\leq M \sum_{k\geq 1} b_{k} - b_{k+1} \\ &\leq 2M(b_1 - b_{n+1})\\ &\leq 2M b_1 .\end{align*}$

theorem (Abel's Theorem):

If $\sum_{k=1}^\infty c_k z^j$ converges on ${\left\lvert {z} \right\rvert} < 1$ then $\begin{align*} \lim_{z\to 1^-} \sum_{k\in {\mathbb{N}}} c_k z^k = \sum_{k\in {\mathbb{N}}} c_k .\end{align*}$

lemma (Abel's Test):

If $f(z) \coloneqq\sum c_k z^k$ is a power series with $c_k \in {\mathbb{R}}^{\geq 0}$ and $c_k\searrow 0$ , then $f$ converges on $S^1$ except possibly at $z=1$ .

example (application of Abel's theorem):

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 $f(1) = \log(2)$ .

remark:

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 $s$ .

proposition (Summation by Parts):

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: $\begin{align*} A_{n} B_{n}=\sum_{k=1}^{n} A_{k} b_{k}+\sum_{k=1}^{n} a_{k} B_{k-1} .\end{align*}$

proof (?):

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

exercise (Stein/Shakarchi 1.20: Series convergence on the circle):

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 $1$ .

solution:

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.

exercise (?):

Show that $\prod_{n\in {\mathbb{Z}}} (1 + a_n) < \infty$ if $\left\{{a_n}\right\} \in \ell_1({\mathbb{Z}})$ .

exercise (Application of summation by parts):

Use summation by parts to show that $\sin(n)/n$ converges.

exercise (?):

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.

#complex/exercise/work

Analytic Number Theory Faves

Exercises