[Paper Review] On an infinite series for $(1+1/x)^x$

Cristinel Mortici, Yue Hu|arXiv (Cornell University)|Jun 30, 2014

Approximation Theory and Sequence Spaces5 references3 citations

TL;DR

This paper constructs an improved asymptotic expansion for $(1+1/x)^x$ using a series in negative powers of $x + \frac{11}{12}$, proving that $\varepsilon = \frac{11}{12}$ yields the fastest convergence among such shifts. It derives a recurrence for coefficients $d_n$, provides an integral representation, and extends Yang's conjecture on Carleman-type inequalities by optimizing the shift parameter.

ABSTRACT

The aim of this paper is to construct a new expansion of $(1+1/x)^x$ related to Carleman's inequality. Our results extend some results of Yang [Approximations for constant e and their applications J. Math. Anal. Appl. 262 (2001) 651-659].

Motivation & Objective

• To resolve an open problem posed by Yang regarding the optimal shift $\varepsilon$ in the expansion $\left(1+\frac{1}{x}\right)^x = e\left(1 - \sum_{n=1}^\infty \frac{d_n}{(x+\varepsilon)^n}\right)$.
• To construct a rapidly converging series expansion for $(1+1/x)^x$ by choosing $\varepsilon = \frac{11}{12}$, which outperforms previous expansions in terms of convergence rate.
• To derive a recurrence relation for the coefficients $d_n$ in the series expansion centered at $x + \frac{11}{12}$.
• To provide an integral representation for the coefficients $d_n$ using the function $g(s) = \frac{1}{\pi} s^s (1-s)^{1-s} \sin(\pi s)$.

Proposed method

• The authors use Lemma 1, which links the limit of $n^k(\omega_n - \omega_{n+1})$ to the rate of convergence of $\omega_n$, to determine the optimal shift $\varepsilon = \frac{11}{12}$.
• They derive a series expansion of the form $\left(1+\frac{1}{n}\right)^n = e\left(1 - \frac{1/2}{n+11/12} - \frac{5/288}{(n+11/12)^3} - \cdots \right)$, with coefficients computed via asymptotic matching.
• A recurrence relation for $d_n$ is established using the generating function $g(t) = \left(\frac{1 - \frac{11}{12}t}{1 + \frac{1}{12}t}\right)^{\frac{11}{12} - \frac{1}{t}}$, which satisfies $g(t) = e(1 + c_1 t + c_2 t^2 + \cdots)$.
• The coefficients $d_n$ are derived from the recurrence $d_n = \frac{1}{n} \sum_{k=0}^{n-1} a_{n-k-1} d_k$ with $d_0 = -1$, where $a_n$ involves powers of 11 and alternating signs.
• An integral representation for $d_n$ is obtained by substituting $t = (x + 11/12)^{-1}$ into the known integral formula for $e - (1+1/x)^x$.
• The integral formula $d_n = \frac{(-1)^n}{12^{n-1}} \left( -\frac{1}{2} + \frac{1}{e} \int_0^1 \frac{(12s-11)^{n-1} - 1}{s-1} g(s) ds \right)$ is derived for $n \geq 2$.

Experimental results

Research questions

• RQ1 What value of $\varepsilon \in (0,1]$ optimizes the convergence rate of the series expansion $\left(1+\frac{1}{x}\right)^x = e\left(1 - \sum_{n=1}^\infty \frac{d_n}{(x+\varepsilon)^n}\right)$?
• RQ2 Can a recurrence relation be derived for the coefficients $d_n$ in the expansion centered at $x + \frac{11}{12}$?
• RQ3 How can the coefficients $d_n$ be represented via an integral involving $g(s) = \frac{1}{\pi} s^s (1-s)^{1-s} \sin(\pi s)$?
• RQ4 Does the series with $\varepsilon = \frac{11}{12}$ yield faster convergence than previous expansions, such as those based on $n+1$?
• RQ5 What is the precise structure of the asymptotic expansion of $(1+1/x)^x$ when the shift is optimized to $\frac{11}{12}$?

Key findings

• The value $\varepsilon = \frac{11}{12}$ is optimal for the series expansion of $(1+1/x)^x$, yielding the fastest convergence rate among all $\varepsilon \in (0,1]$.
• The series $\left(1+\frac{1}{n}\right)^n = e\left(1 - \frac{1/2}{n+11/12} - \frac{5/288}{(n+11/12)^3} - \cdots \right)$ has a truncation error of order $n^{-k}$ after $k$ terms, outperforming the $n^{-(k-1)}$ error of previous expansions.
• The coefficients $d_n$ satisfy the recurrence $d_n = \frac{1}{n} \sum_{k=0}^{n-1} a_{n-k-1} d_k$ with $d_0 = -1$, where $a_n = \frac{n+1}{12^{n+2}} \left( \frac{(-1)^{n+1}11 - 11^{n+2}}{n+1} - \frac{(-1)^n - 11^{n+2}}{n+2} \right)$.
• An integral representation for $d_n$ is given by $d_n = \frac{(-1)^n}{12^{n-1}} \left( -\frac{1}{2} + \frac{1}{e} \int_0^1 \frac{(12s-11)^{n-1} - 1}{s-1} g(s) ds \right)$ for $n \geq 2$, where $g(s) = \frac{1}{\pi} s^s (1-s)^{1-s} \sin(\pi s)$.
• The first few coefficients are $d_1 = \frac{1}{2}$, $d_2 = 0$, $d_3 = \frac{5}{288}$, $d_4 = \frac{139}{17280}$, and $d_5 = \frac{119}{23040}$, confirming the improved convergence.