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I am trying to analyze the sequence $g(n)$, defined as follows:

  • $g(1)=0$
  • $g(2)=1$
  • For $n\geq 2$, $g(n+1)=g(n)\left(1+\displaystyle{\frac{n}{n-1}\ln\left(\frac{n+1}{n}\right)}\right)$.

A computational experiment suggests that, as $n\rightarrow\infty$, $\displaystyle{\frac{g(n)}{n}}$ approaches a constant of $\approx 0.758097$, but I don't know how to prove it. How would you do so?

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3 Answers 3

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I shall derive a complete asymptotic expansion. Let $$ C := \prod\limits_{k = 2}^\infty {\left( {1 - \frac{1}{k} + \log \left( {1 + \frac{1}{k}} \right)} \right)} = 0.758096959930697113026\ldots $$ and $$ f(n) := \left[ {\prod\limits_{k = n}^\infty {\left( {1 - \frac{1}{k} + \log \left( {1 + \frac{1}{k}} \right)} \right)} } \right]^{ - 1} $$ for any $n\ge 1$. Then, for $n \ge 3$, \begin{align*} g(n) & = \prod\limits_{k = 2}^{n - 1} {\left( {1 + \frac{k}{{k - 1}}\log \left( {\frac{{k + 1}}{k}} \right)} \right)} = \prod\limits_{k = 2}^{n - 1} {\frac{k}{{k - 1}}\left( {1 - \frac{1}{k} + \log \left( {1 + \frac{1}{k}} \right)} \right)} \\ & = (n-1)\prod\limits_{k = 2}^{n - 1} {\left( {1 - \frac{1}{k} + \log \left( {1 + \frac{1}{k}} \right)} \right)} = C \cdot (n-1) \cdot f(n). \end{align*} Now \begin{align*} f(n) & = \exp \left( { - \sum\limits_{k = n}^\infty {\log\left( {1 - \frac{1}{k} + \log \left( {1 + \frac{1}{k}} \right)} \right)} } \right) = \exp \left( {\sum\limits_{k = n}^\infty {\sum\limits_{m = 2}^\infty {(-1)^m\frac{{a_m }}{{m!}}\frac{1}{{k^m }}} } } \right) \\ & = \exp \left( {\sum\limits_{m = 2}^\infty {(-1)^m\frac{{a_m }}{{m!}}\sum\limits_{k = n}^\infty {\frac{1}{{k^m }}} } } \right) = \exp \left( {\sum\limits_{m = 2}^\infty {(-1)^m\frac{{a_m }}{{m!}}\zeta (m,n)} } \right), \end{align*} where the sequence $a_m$ is $\mathrm{A}331559$ in the OEIS, and $\zeta(m,n)$ is the Hurwitz zeta function. Now for each fixed $m$, $$ \zeta (m,n) \sim \frac{1}{{m - 1}}\frac{1}{{n^{m - 1} }} + \frac{1}{2}\frac{1}{{n^m }} + \sum\limits_{k = 2}^\infty {\frac{{B_k }}{{k!}}\frac{{(m)_{k - 1} }}{{n^{m + k - 1} }}} $$ as $n\to +\infty$ (cf. $(25.11.43)$). Here $B_k$ denotes the Bernoulli numbers and $(m)_k$ is the Pochhammer symbol. Substitution and re-arrangement then gives $$ \sum\limits_{m = 2}^\infty {(-1)^m\frac{{a_m }}{{m!}}\zeta (m,n)} \sim \sum\limits_{m = 1}^\infty {\frac{{b_m }}{{n^m }}} $$ as $n\to +\infty$, where $b_1 = \frac{1}{2}$, $b_2 = \frac{1}{12}$ and $$ b_m = (-1)^{m+1}\frac{{a_{m + 1} }}{{m(m + 1)!}} + (-1)^{m}\frac{1}{2}\frac{{a_m }}{{m!}} + \sum\limits_{k = 2}^{m - 1} {(-1)^{m-k+1}\frac{{B_k \, a_{m - k + 1} }}{{k!(m - k + 1)!}}(m - k + 1)_{k - 1} } , $$ for $m\ge 3$. Exponentiation of this asymptotic expansion then yields $$ g(n) \sim C \cdot (n-1) \cdot \sum\limits_{m = 0}^\infty {\frac{{c_m }}{{n^m }}} = C \cdot (n-1)\left( {1 + \frac{1}{{2n}} + \frac{5}{{24n^2 }} + \frac{5}{{48n^3 }} + \frac{{287}}{{5760n^4 }} + \ldots } \right) $$ as $n\to +\infty$, where $c_0=1$ and $$ c_m = \frac{1}{m}\sum\limits_{k = 1}^m {kb_k c_{m - k} } $$ for $m\ge 1$. A more natural form is $$ g(n) \sim C \cdot n \cdot \sum\limits_{m = 0}^\infty {\frac{{d_m }}{{n^m }}} = C \cdot n\left( {1 - \frac{1}{{2n}} - \frac{7}{{24n^2 }} - \frac{5}{{48n^3 }} - \frac{{313}}{{5760n^4 }} + \ldots } \right) $$ as $n\to +\infty$, where $d_0=1$ and $$ d_m = c_m - c_{m - 1} $$ for $m\ge 1$.

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Define $\displaystyle h(n) = 1-\frac1n+\ln\biggl( 1+\frac1n \biggr)$, so that \begin{align*} g(n+1) &= \prod_{k=2}^n \biggl( 1+\frac k{k-1}\ln\frac{k+1}k \biggr) \\ &= \prod_{k=2}^n \biggl( \frac k{k-1}h(k) \biggr) = \prod_{k=2}^n \frac k{k-1} \prod_{k=2}^n h(k) = n \prod_{k=2}^n h(k). \end{align*} But from the Maclaurin series for $\ln(1+x)$, we see that $h(n) \displaystyle = 1+O\biggl( \frac1{n^2} \biggr)$, and so this last product converges as $n\to\infty$. Indeed this equation implies $$ g(n) = n \prod_{k=2}^\infty h(k)+O(1) $$ where this infinite product is indeed approximately $0.75809696$, accurate to that many decimal places (rounded).

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    $\begingroup$ If $C$ denotes your constant then $$ g(n) \sim Cn\left( {1 - \frac{1}{{2n}} - \frac{7}{{24n^2 }} - \frac{5}{{48n^3 }} - \frac{{313}}{{5760n^4 }} + \ldots } \right) $$ as $n\to +\infty$. $\endgroup$
    – Gary
    Mar 21, 2023 at 7:45
  • $\begingroup$ @Gary could you explain how you got these? Are there some standard tools to find this? $\endgroup$ Mar 21, 2023 at 14:57
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    $\begingroup$ @DiegoSantos I posted an answer. $\endgroup$
    – Gary
    Mar 22, 2023 at 6:39
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Some loose bounds: $$\dfrac{g(n+1)}{g(2)} = \prod_{k=2}^n\dfrac{g(k+1)}{g(k)} = \prod_{k=2}^n\left(1 + \dfrac{k}{k-1}\ln\frac{k+1}{k}\right)< \prod_{k=2}^n\left(1+\frac 1k\right)^{\frac{k}{k-1}}$$ and also: $$g(n+1)<g(2)\prod_{k=2}^n\left(1 + \dfrac{1}{k-1}\right) = n$$ by using $\ln(1+x)<x$ and by using $\dfrac{x}{1+x}<\ln x:$ $$g(n+1) > \prod_{k=2}^n\left(1 + \dfrac{k}{k^2-1}\right)>\prod_{k=2}^n\left(1 + \dfrac{1}{k}\right) = \dfrac{n+1}{2}.$$ I don't know if the infinite products have closed form in the limit though.

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