A tricky convergence problem

This is a slight simplification of the Math.SE user @Mike's wonderful solution [1] that is rephrased in my own words.

Theorem. Suppose

1. $(a_n)_{n\geq 1}$ is a non-negative, non-increasing sequence,
2. $(f(n))_{n\geq 1}$ is a sequence of positive integers such that $f(n) \to \infty$, and
3. $\sum_{n\geq 1} a_{f(n)} $ converges.

Then $\sum_{n\geq 1} \frac{a_n}{f(n)}$ also converges.

Proof. We partition $\mathbb{N}_1 = \{1, 2, \ldots\}$ into two subsets $A$ and $B$, where \begin{align*} A &= \{ n \in \mathbb{N}_1 : f(n) > n \}, \\ B &= \{ n \in \mathbb{N}_1 : f(n) \leq n \}. \end{align*} Then it suffices to show that both $ \sum_{n\in A} \frac{a_n}{f(n)} $ and $ \sum_{n\in B} \frac{a_n}{f(n)} $ are finite.

First sum. Since $(a_n)_{n\in\mathbb{N}_1}$ is non-increasing and $1 \leq f(n) \leq n$ for each $n \in B$, we get $$ \sum_{n\in B} \frac{a_n}{f(n)} \leq \sum_{n \in B} a_{f(n)} \lt \infty. $$

Second sum. Now enumerate $f(\mathbb{N}_1)$ as $\{k_1 \lt k_2 \lt \ldots \}$, and define $$A_i = A \cap [k_i, k_{i+1})$$ for each $i$. Then the following implication holds true: \begin{align*} n \in A_i \quad\implies\quad f(n) \gt n \geq k_i \quad\implies\quad f(n) \geq k_{i+1}. \end{align*} So it follows that $$ \sum_{n \in A_i} \frac{a_n}{f(n)} \leq \sum_{n \in A_i} \frac{a_{k_i}}{k_{i+1}} = \frac{k_{i+1} - k_i}{k_{i+1}} a_{k_i} \leq a_{k_i}. $$ Summing both sides over $i$, \begin{align*} \sum_{n \in A} \frac{a_n}{f(n)} &\leq \sum_{i=1}^{\infty} a_{k_i} \\ &\leq \sum_{i=1}^{\infty} \sum_{n \mathbin{:} f(n) = k_i} a_{f(n)} \\ &= \sum_{n=1}^{\infty} a_{f(n)} \lt \infty. \end{align*} Therefore the desired conclusion follows.

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References.

1. [1] Mike, If $\sum_{n=1}^{\infty}{a_{[f(n)]}}$ converges, then $\sum_{n=1}^{\infty}{\frac{a_n}{f(n)}}$ converges., URL (version: 2022-06-05): https://math.stackexchange.com/q/4464970

A property of an integrable function on $\mathbb{R}$

Theorem. Let $f \in L^1(\mathbb{R}, \mathrm{d}x)$. Then $$ \lim_{n\to\infty} f(nx) = 0 \quad \text{for a.e. } x. $$

The proof below is essentially what is shown in [1], written in my own words.

Proof. Fix $\varepsilon \gt 0$ and define $E = \{x \in \mathbb{R} : |f(x)| \geq \varepsilon\}$. Then for any $[a, b] \subseteq (0, \infty)$, \begin{align*} \int_{a}^{b} \sum_{n=1}^{\infty} \mathbf{1}_{E}(nx) \, \mathrm{d}x &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{na}^{nb} \mathbf{1}_{E}(t) \, \mathrm{d}t \\ &= \int_{E} \sum_{n=1}^{\infty} \frac{1}{n} \mathbf{1}_{\{na \leq t \leq nb\}} \, \mathrm{d}t \\ &= \int_{E} \sum_{n=1}^{\infty} \frac{1}{n} \mathbf{1}_{\{ t/b \leq n \leq t/a \}} \, \mathrm{d}t \\ &\leq \int_{E} (2 + \log(b/a)) \, \mathrm{d}t \\ &= (2 + \log(b/a)) |E|, \end{align*} where $|E|$ denotes the Lebesgue measure of $E$. Since $f$ is integrable, $|E| \lt \infty$. So the sum $\sum_{n=1}^{\infty} \mathbf{1}_{E}(nx)$ is finite and hence $nx \in E$ for only finitely many $n$ for a.e. $x$ in $[a, b]$. Since $[a, b]$ is arbitrary, the same is true for a.e. $x$ in $\mathbb{R}$. From this, it is now routine to conclude the desired claim. $\square$

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References

• [1] Emmanuel Lesigne. On the behavior at infinity of an integrable function. American Mathematical Monthly, Mathematical Association of America, 2010, 117 (2), pp.175-181. ⟨hal-00276738v3⟩

A simple but useful estimate

Theorem. Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous, integrable function of bounded variation. Then $\sum_{n\in\mathbb{Z}} f(n)$ converges absolutely, and we have $$ \left| \sum_{n\in \mathbb{Z}} f(n) - \int_{\mathbb{R}} f(x) \, \mathrm{d}x \right| \leq \frac{1}{2} V(f), $$ where $V(f) = \int_{\mathbb{R}} |\mathrm{d}f(x)|$ denotes the total variation of $f$ over $\mathbb{R}$.

Proof. Let $\tilde{B}(x) = x - \lfloor x \rfloor - \frac{1}{2}$. Then performing integration by parts, for $a \le b$, \begin{align*} \int_{a}^{b} f(x) \, \mathrm{d}x - \sum_{n \in [a, b]\cap\mathbb{Z}} f(n) &= \int_{[a, b]} f(x) \, \mathrm{d}\tilde{B}(x) \\ &= [f(x)\tilde{B}(x)]_{a^-}^{b} - \int_{[a,b]} \tilde{B}(x) \, \mathrm{d}f(x). \end{align*} Using the fact that $|\tilde{B}(x)| \leq \frac{1}{2}$, it therefore follows that \begin{align*} \left| \int_{a}^{b} f(x) \, \mathrm{d}x - \sum_{n \in [a, b]\cap\mathbb{Z}} f(n) \right| \leq \frac{|f(a)| + |f(b)|}{2} + \frac{1}{2} \int_{[a,b]} \, |\mathrm{d}f(x)| \end{align*} From the assumption, it is easy to check that $f(x) \to 0$ as $|x| \to \infty$. So, the above estimate shows that $\sum_{n\in\mathbb{Z}} f(n)$ satisfies the Cauchy criterion and hence converges as $a \to -\infty$ and $b \to \infty$. Moreover, noting that $|f|$ also satisfies the assumption in plcae of $f$, we find that this sum converges absolutely. Finally, passing to the limit as $[a,b] \uparrow \mathbb{R}$ proves the desired bound. $\square$

Cauchy–Binet formula

Write $[n]$ for the set $\{1,2,\ldots,n\}$ and $\binom{[n]}{k}$ for the family of all subets of $[n]$ and size $k$. Also, for an $m \times n$ matrix $X$, $I \subseteq [m]$, and $J \subseteq [n]$, let $X_{I,J}$ denote the submatrix obtained from $X$ by deleting row $i$ for each $i \notin I$ and column $j$ for each $j\notin [J]$.

Theorem. (Cauchy–Binet formula) Let $A$ be a $k\times n$ matrix and $B$ be an $n\times k$ matrix over the field $F$. Then $$\det(AB) = \sum_{S\subseteq\binom{[n]}{k}} \det(A_{[k],S}B_{S,[k]}).$$

1^st Proof. Let $(e_1,\ldots, e_m)$ denote the standard basis of $F^m$. (For convenience, we will abuse the notation so $e_i$ stands for the $i$th standard basis vector in any of $F^m$, $m \geq i$.) Then \begin{align*} \det(AB)(e_1 \wedge \cdots \wedge e_k) &= \bigl( {\textstyle\bigwedge}^k AB \bigr)(e_1 \wedge \cdots \wedge e_k) \\ &= \bigl( {\textstyle\bigwedge}^k A \bigr)[(Be_1) \wedge \cdots \wedge (Be_k)]. \end{align*} Now by noting that $Be_j = \sum_{i=1}^{n} B_{ij} e_i$ is the $j$th column of $B$, we may expand the exterior product in the last line as \begin{align*} &[(Be_1) \wedge \cdots \wedge (Be_k)] \\ &= \sum_{\substack{\rho : [k] \to [n] \\ \sigma \text{ injective} }} (B_{\rho(1),1} e_{\rho(1)}) \wedge \cdots \wedge (B_{\rho(k),k} e_{\rho(k)} ) \\ &= \sum_{\substack{\sigma : [k] \to [n] \\ \sigma \text{ increasing} }} \sum_{\substack{\tau : [k] \to [k] \\ \tau \text{ bijective}}} \left( \prod_{i=1}^{k} B_{\sigma(\tau(i)),i} \right) (e_{\sigma(\tau(1))} \wedge \cdots \wedge e_{\sigma(\tau(k))} ) \\ &= \sum_{\substack{\sigma : [k] \to [n] \\ \sigma \text{ increasing} }} \sum_{\substack{\tau : [k] \to [k] \\ \tau \text{ bijective}}} \operatorname{sgn}(\tau) \left( \prod_{i=1}^{k} B_{\sigma(\tau(i)),i} \right) (e_{\sigma(1)} \wedge \cdots \wedge e_{\sigma(k)} ) \\ &= \sum_{\substack{\sigma : [k] \to [n] \\ \sigma \text{ increasing} }} \det(B_{\sigma([k]),[k]}) (e_{\sigma(1)} \wedge \cdots \wedge e_{\sigma(k)} ) \end{align*} Plugginb this back, we therefore get \begin{align*} &\det(AB)(e_1 \wedge \cdots \wedge e_k) \\ &= \sum_{\substack{\sigma : [k] \to [n] \\ \sigma \text{ increasing} }} \det(B_{\sigma([k]),[k]}) [(Ae_{\sigma(1)}) \wedge \cdots \wedge (Ae_{\sigma(k)}) ] \\ &= \sum_{\substack{\sigma : [k] \to [n] \\ \sigma \text{ increasing} }} \det(B_{\sigma([k]),[k]}) \det(A_{[k],\sigma([k])}) (e_1 \wedge \cdots \wedge e_k) \\ &= \sum_{S \in \binom{[n]}{k}} \det(A_{[k],S}) \det(B_{S,[k]}) (e_1 \wedge \cdots \wedge e_k). \end{align*} This completes the proof. $\square$

2^nd Proof. We specialize in the case $F = \mathbb{C}$. Also, for each index set $I$, we abbreviate $\wedge_{i\in I} e_i = \wedge I$. Then using inner product between multivectors, we get \begin{align*} \det(AB) &= \left\langle (\wedge^k A^*)(\wedge [k]), (\wedge^k B)(\wedge [k]) \right\rangle \\ &= \sum_{S \in \binom{[n]}{k}} \left\langle (\wedge^k A^*)(\wedge [k]), \wedge S \right\rangle \left\langle \wedge S, (\wedge^k B)(\wedge [k]) \right\rangle \\ &= \sum_{S \in \binom{[n]}{k}} \det(A_{[k],S})\det(B_{S,[k]}) \end{align*}

A similar technique shows:

Proposition. (Coefficients of a characteristic Polynomial) Let $A$ be an $n\times n$ matrix. Then $$\det(zI_n + A) = \sum_{k=0}^{n} z^{n-k} \sum_{S\subseteq\binom{[n]}{k}} \det(A_{S,S}).$$

Proposition. Let $a_1, \ldots, a_p \in \mathbb{C}$ and $v_1, \ldots, v_p \in \mathbb{C}^n$. Then $$\det\left( \sum_{j=1}^{p} a_j v_j v_j^{*} \right) = \sum_{\{j_1 \lt \ldots \lt j_n \} \subseteq [p]} (a_{j_1} \cdots a_{j_n}) \left| \det( v_{j_1}, \ldots, v_{j_n} ) \right|^2. $$

Determinant of a Cauchy matrix

Throughout this posting, the expressions $$ \sum_{\mathit{vars} \mathop{:} \mathit{conds}} f(\mathit{vars}) \qquad\text{and}\qquad \prod_{\mathit{vars} \mathop{:} \mathit{conds}} f(\mathit{vars}) $$ will refer to the sum and product of $f(\mathit{vars})$ over all possible values of $\mathit{vars}$ subject to the constraints $\mathit{conds}$, respectively.

Theorem. (Determinant of a Cauchy matrix) Let $A$ be an $n\times n$ matrix with elements $a_{ij}$ in the form $$ a_{ij} = \frac{1}{x_i + y_j}. $$ (Think of $A$ as an matrix over the field of rational functions in variables $x_i$'s and $y_j$'s.) Then $$ \det A = \frac{\prod_{i, j \mathbin{:} i \lt j} (x_i - x_j)(y_i - y_j)}{\prod_{i, j} (x_i + y_j)}. $$

Proof. We prove the claim by induction on $n$. The base case is clearly true, so it suffices to establish the inductive step.

Suppose that the claim holds for all $(n-1)\times(n-1)$ matrices. Also, recall that if $p(z)$ is a polynomial of degree less than $n$ and if $\alpha_1, \ldots, \alpha_n$ are distinct, then we have the partial fraction decomposition $$ \frac{p(z)}{\prod_{j} (z - \alpha_j)} = \sum_{k} \frac{p(\alpha_k)}{(z - \alpha_k) \prod_{j \mathbin{:} j \neq k} (\alpha_k - \alpha_j)}. $$ So, by regarding $x_n$ as the variable and applying the partial fraction decomposition, \begin{align*} &\frac{\prod_{i \mathbin{:} i \lt n} (x_i - x_n)}{\prod_{j} (x_n + y_j)} \\ &= \sum_{k} \frac{\prod_{i \mathbin{:} i \lt n} (x_i + y_k)}{(x_n + y_k) \prod_{j \mathbin{:} j \neq k} (y_j - y_k)} \\ &= \sum_{k} (-1)^{n-k} \frac{\prod_{i \mathbin{:} i \lt n} (x_i + y_k)}{(x_n + y_k) \bigl[ \prod_{j \mathbin{:} j \lt k} (y_j - y_k) \bigr]\bigl[ \prod_{j \mathbin{:} k \lt j} (y_k - y_j) \bigr]} \end{align*} From this, it follows that \begin{align*} &\frac{\prod_{i, j \mathbin{:} i \lt j} (x_i - x_j)(y_i - y_j)}{\prod_{i, j} (x_i + y_j)} \\ &= \frac{\bigl[ \prod_{i, j \mathbin{:} i \lt j \lt n} (x_i - x_j) \bigr]\bigl[ \prod_{i, j \mathbin{:} i \lt j} (y_i - y_j) \bigr]}{\prod_{i, j \mathbin{:} i \lt n} (x_i + y_j)} \cdot \frac{\prod_{i \mathbin{:} i \lt n} (x_i - x_n)}{\prod_{j} (x_n + y_j)} \\ &= \sum_{k} \frac{(-1)^{n-k}}{x_n + y_k} \cdot \frac{\bigl[ \prod_{i, j \mathbin{:} i \lt j \lt n} (x_i - x_j) \bigr]\bigl[ \prod_{i, j \mathbin{:} i \lt j; i \neq k; j \neq k} (y_i - y_j) \bigr]}{\prod_{i, j \mathbin{:} i \lt n; j\neq k} (x_i + y_j)} \end{align*} However, the last line is precisely the cofactor expansion of $\det A$ along the $n$th row, where the determinants of the submatrices are computed using the inductive hypothesis. Therefore the inductive step is established and the proof is done. $\square$