Using Cauchy condensation test to derive the results about generalized Bertrand series

Question

The generalized Bertrand series is defined by:

$$B_{k}:=\sum_{n\ge N}^{ }\frac{1}{n\log n\log\log n\cdot\cdot\cdot\log^{\circ\left(k-1\right)}n\cdot\left(\log^{\circ k}n\right)^{p}},$$

where $N= \left \lfloor{e^{\circ k}\left(0\right)}\right \rfloor +1$, $k \in \mathbb N^{+}$ and $p$ non-negative number.

Theorem: The series converges for $p>1$ and diverges for $0<p\le1$.

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I want to know is it possible to use Cauchy condensation test to derive the same result? since it's been mentioned here, (Of course using Schlömilch's Generalization is also acceptable).

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Answer 1

$\def\d{\mathrm{d}}$Here is an approach using telescoping. Since$$ (\ln^{\circ k} x)' = \left( \prod_{j = 0}^{k - 1} \ln^{\circ j} x \right)^{-1} $$ by the chain rules, then$$ \int \left( (\ln^{\circ k} x)^p \prod_{j = 0}^{k - 1} \ln^{\circ j} x \right)^{-1} \,\d x = \int \frac{\d(\ln^{\circ k} x)}{(\ln^{\circ k} x)^p} =\begin{cases} -\dfrac{p - 1}{(\ln^{\circ k} x)^{p - 1}}; & p ≠ 1\\ \ln^{\circ k + 1} x; & p = 1 \end{cases}. $$ Define $f_p(x) := \left( (\ln^{\circ k} x)^p \prod\limits_{j = 0}^{k - 1} \ln^{\circ j} x \right)^{-1}$. Note that no matter $p \geqslant 0$ or $p < 0$, there exists an integer $M_p \geqslant N := [\mathrm{e}^{\circ k}(0)] + 1$ such that $f_p$ is decreasing on $[M_p, +∞)$.

Case 1: $p < 1$. For $n \geqslant M_p$,$$ f_p(n) = \int_n^{n + 1} f_p(n) \,\d x \geqslant \int_n^{n + 1} f_p(x) \,\d x = (1 - p) ((\ln^{\circ k} (n + 1))^{1 - p} - (\ln^{\circ k} n)^{1 - p}), $$ thus\begin{gather*} \sum_{n = N_p}^∞ f_p(n) \geqslant \sum_{n = M_p}^∞ f_p(n) \geqslant (1 - p) \sum_{n = M_p}^∞ ((\ln^{\circ k} (n + 1))^{1 - p} - (\ln^{\circ k} n)^{1 - p})\\ = (1 - p) \lim_{n → ∞} ((\ln^{\circ k} (n + 1))^{1 - p} - (\ln^{\circ k} M_p)^{1 - p}) = +∞. \end{gather*}

Case 2: $p = 1$. For $n \geqslant M_p$,$$ f_p(n) = \int_n^{n + 1} f_p(n) \,\d x \geqslant \int_n^{n + 1} f_p(x) \,\d x = \ln^{\circ k + 1} (n + 1) - \ln^{\circ k + 1} n, $$ and $\sum\limits_{n = N_p}^∞ f_p(n)$ diverges analogously.

Case 3: $p > 1$. For $n \geqslant M_p + 1$,$$ f_p(n) = \int_{n - 1}^n f_p(n) \,\d x \leqslant \int_{n - 1}^n f_p(x) \,\d x = (p - 1) \left( \frac{1}{(\ln^{\circ k} n)^{p - 1}} - \frac{1}{(\ln^{\circ k} (n + 1))^{p - 1}} \right), $$ thus\begin{align*} &\mathrel{\phantom{=}}{} \sum_{n = N_p}^∞ f_p(n) = \sum_{n = N_p}^{M_p} f_p(n) + \sum_{n = M_p}^∞ f_p(n)\\ &\leqslant \sum_{n = N_p}^{M_p} f_p(n) + (p - 1) \sum_{n = M_p}^∞ \left( \frac{1}{(\ln^{\circ k} n)^{p - 1}} - \frac{1}{(\ln^{\circ k} (n + 1))^{p - 1}} \right)\\ &= \sum_{n = N_p}^{M_p} f_p(n) + \frac{p - 1}{(\ln^{\circ k} M_p)^{p - 1}} < +∞. \end{align*}

Answer 2

Here is an approach using the condensation test.

Lemma 1: If $\{x_n\}$ and $\{y_n\}$ are sequences of positive numbers such that $y_n \geqslant M$ for some constant $M > 1$ and $\lim\limits_{n → ∞} \dfrac{x_n}{y_n} = 1$, then $\lim\limits_{n → ∞} \dfrac{\ln x_n}{\ln y_n} = 1$.

Lemma 2: For any integer $k \geqslant 2$,$$ \lim_{n → ∞} \frac{\ln^{\circ k}(2^n)}{\ln^{\circ k - 1} n} = 1. $$

(Lemma 2 can be proved by induction on $k$ with Lemma 1.)

Now return to the question and prove by induction on $k$. For $k = 0$, it is well-known that$$ \sum \frac{1}{n^p} < +∞ \Longleftrightarrow p > 1. $$ For $k = 1$, note that $\dfrac{1}{2^n (\ln(2^n))^p} = \dfrac{1}{(\ln 2)^p · 2^n n^p}$, thus by the condensation test,$$ \sum \frac{1}{n (\ln n)^p} < +∞ \Longleftrightarrow \sum 2^n · \frac{1}{2^n (\ln(2^n))^p} < +∞ \Longleftrightarrow \sum \frac{1}{n^p} < +∞ \Longleftrightarrow p > 1. $$ Now assume that$$ \sum \left( (\ln^{\circ k} n)^p \prod_{j = 0}^{k - 1} \ln^{\circ j} n \right)^{-1} < +∞ \Longleftrightarrow p > 1 $$ for some $k \geqslant 1$. Note that by Lemma 2,\begin{gather*} (\ln^{\circ k + 1}(2^n))^p \prod_{j = 0}^k \ln^{\circ j}(2^n) = 2^n · n \ln 2 · (\ln^{\circ k + 1}(2^n))^p \prod_{j = 2}^k \ln^{\circ j}(2^n)\\ \sim 2^n · n \ln 2 · (\ln^{\circ k} n)^p \prod_{j = 2}^k \ln^{\circ j - 1} n = 2^n \ln 2 · (\ln^{\circ k} n)^p \prod_{j = 0}^{k - 1} \ln^{\circ j} n \end{gather*} as $n → ∞$, thus by the condensation test,\begin{gather*} \sum \left( (\ln^{\circ k + 1} n)^p \prod_{j = 0}^k \ln^{\circ j} n \right)^{-1} < +∞ \Longleftrightarrow \sum 2^n · \left( (\ln^{\circ k + 1}(2^n))^p \prod_{j = 0}^k \ln^{\circ j}(2^n) \right)^{-1} < +∞\\ \Longleftrightarrow \sum \left( (\ln^{\circ k} n)^p \prod_{j = 0}^{k - 1} \ln^{\circ j} n \right)^{-1} < +∞ \Longleftrightarrow p > 1. \end{gather*} End of induction.

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Proof of Lemma 2: For $k = 2$,$$ \lim_{n → ∞} \frac{\ln(\ln(2^n))}{\ln n} = \lim_{n → ∞} \frac{\ln(n\ln 2)}{\ln n} = \lim_{n → ∞} \frac{\ln n + \ln(\ln 2)}{\ln n} = 1 + \lim_{n → ∞} \frac{\ln(\ln 2)}{\ln n} = 1. $$ Assume the proposition holds for $k$, then by Lemma 1,$$ \lim_{n → ∞} \frac{\ln^{\circ k + 1}(2^n)}{\ln^{\circ k} n} = \lim_{n → ∞} \frac{\ln(\ln^{\circ k}(2^n))}{\ln(\ln^{\circ k - 1} n)} = 1. $$ End of induction.