# Asymptotic behavior of sequence

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?

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 $$m\ge 1$$.

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).

• 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 $n\to +\infty$.
– Gary
Mar 21, 2023 at 7:45
• @Gary could you explain how you got these? Are there some standard tools to find this? Mar 21, 2023 at 14:57
• @DiegoSantos I posted an answer.
– Gary
Mar 22, 2023 at 6:39

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: $$g(n+1) by using $$\ln(1+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.