# Euler常数γ的推广

Xiu-You Xu*

Email:xiuyouxu#gmail.com

## abstract

When $n\rightarrow\infty$, it is well known that $\sum\nolimits_{i=1}^n\frac{1}{i}\sim{\ln{n}+\gamma}$, where $\gamma$ is Euler-Mascheroni Constant. We will talk about the coefficients of the function having $(-lnx)^m$ as $n$th-order derivative, and generalize $\gamma$ to a sequence $\{\gamma_m\}$ and get the asymptotic expansion of a sequence $a_n^i$ as $n\rightarrow\infty$, that is $a_n^i=\sum\nolimits_{j=1}^n(-1)^{j-1}C_n^jj^{-i}\sim\sum\nolimits_{j=0}^i\frac{\gamma_{i-j}}{j!}(\ln{n})^j$.

## Introduction

We will talk about the coefficients of integral of function of natural logrithm of power $m$, where $m$ is an integer.

Set $I_n(x)=\frac{x^n}{n!}(\ln{x}-1-\cdots-\frac{1}{n})$, so

\begin{align*}
I_n^{‘}(x)&=\frac{x^{n-1}}{(n-1)!}(\ln{x}-1-\cdots-\frac{1}{n})+\frac{x^n}{n!}\frac{1}{x}\\
&=\frac{x^{n-1}}{(n-1)!}(\ln{x}-1-\cdots-\frac{1}{n-1})\\
&=I_{n-1}(x).
\end{align*}

So the $n$th-order derivative of $I_n(x)$ is $\ln{x}$.

Now we think about the integral of $(-\ln{x})^m$, $m\in{N}$. Set

\begin{align*}
I_{m,1}(x)&=\int{(-\ln{x})^m}dx=x(-\ln{x})^m-\int{xd[(-\ln{x})^m]}\\
&=x(-\ln{x})^m+m\int{(-\ln{x})^{m-1}}dx\\
&=x(-\ln{x})^m+mI_{m-1,1}(x)\\
&=x(-\ln{x})^m+mx(-\ln{x})^{m-1}+m(m-1)I_{m-2,1}(x)\\
&=x(-\ln{x})^m+mx(-\ln{x})^{m-1}+\cdots+m(m-1)\cdots2*I_{1,1}(x)\\
&=m!x[1+\frac{-\ln{x}}{1!}+\frac{(-\ln{x})^2}{2!}+\cdots+\frac{(-\ln{x})^m}{m!}]\\
&=m!x\sum_{i=0}^m\frac{(-\ln{x})^i}{i!}.
\end{align*}

Like $m=1$, we think about the function having $(-\ln{x})^m$ as $n$th-order derivative and set the function is $I_{m,n}(x)$. From $I_{m,1}(x)$ we know that the integral of $(-\ln{x})^m$ is a product of $m!x$ and an $m$-degree polynomial of $\ln{x}$. It is easy to know that $I_{m,n}(x)$ is a product of $\frac{m!x^n}{n!}$ and an $m$-degree polynomial of $\ln{x}$.

Set $I_{m,n}(x)=\frac{m!x^n}{n!}\sum_{i=0}^m\frac{a_n^i(-\ln{x})^{m-i}}{(m-i)!}$, so the derivative of $I_{m,n}(x)$ is

\begin{align*}
I_{m,n}^{‘}(x)&=\frac{m!x^{n-1}}{(n-1)!}\sum_{i=0}^m\frac{a_n^i(-\ln{x})^{m-i}}{(m-i)!}+\frac{m!x^n}{n!}\sum_{i=0}^{m-1}\frac{a_n^i(-\ln{x})^{m-i-1}}{(m-i-1)!}(-\frac{1}{x})\\
&=\frac{m!x^{n-1}}{(n-1)!}[\sum_{i=1}^m\frac{(a_n^i-\frac{a_n^{i-1}}{n})(-\ln{x})^{m-i}}{(m-i)!}+\frac{a_n^0}{m!}(-\ln{x})^m],
\end{align*}

if $I_{m,n}^{‘}(x)=I_{m,n-1}(x)$, we get $a_n^i-\frac{a_n^{i-1}}{n}=a_{n-1}^i(1\le{i}\le{m})$, $a_n^0=a_{n-1}^0$, that is

$a_n^i=a_{n-1}^i+\frac{a_n^{i-1}}{n}, a_n^0=1, n\ge{1}.(1)$

For fixed $n$, set $E_n(x)$ the generating function of $a_n^i$, from equation (1) we have $a_n^ix^i-a_{n-1}^ix^i=\frac{a_n^{i-1}x^i}{n}(n\ge{2})$, sum of the equation from $i=1$ to $\infty$ gives

$E_n(x)-1-[E_{n-1}(x)-1]=\frac{x}{n}E_n(x),$

that is $E_n(x)=\frac{nE_{n-1}(x)}{n-x}$. Because $a_1^i=1$ for all $i\ge{0}$, we have $E_1(x)=\frac{1}{1-x}$, so

$E_n(x)=\frac{n!}{(1-x)(2-x)\cdots(n-x)}.$

Sum of partial fraction of $E_n(x)$ is

$E_n(x)=n![\frac{\alpha_1}{1-x}+\frac{-\alpha_2}{2-x}+\cdots+\frac{(-1)^{n-1}\alpha_n}{n-x}],$

with both sides multiplying by $(j-x)$ and setting $x=j$ gives

$left=\frac{n!}{(-1)^{j-1}(j-1)!(n-j)!}=(-1)^{j-1}n!\alpha_j=right,$

so $\alpha_j=\frac{1}{(j-1)!(n-j)!}=\frac{C_{n-1}^{j-1}}{(n-1)!}$. Think about the power series expansion of $E_n(x)$ of its partial fraction sum, the coefficient of $x^i$ is

$a_n^i=n!\sum_{j=1}^n\frac{(-1)^{j-1}\alpha_j}{j}\frac{1}{j^i}=n!\sum_{j=1}^n(-1)^{j-1}\frac{C_{n-1}^{j-1}j^{-(i+1)}}{(n-1)!}=\sum_{j=1}^n(-1)^{j-1}C_n^jj^{-i},$

we get $a_n^i=\sum_{j=1}^n(-1)^{j-1}C_n^jj^{-i}$.

Here are some values of $a_n^i$:

 n\i 0 1 2 3 $\cdots$ 1 1 1 1 1 $\cdots$ 2 1 3/2 7/4 15/8 $\cdots$ 3 1 11/6 85/36 575/216 $\cdots$ 4 1 25/12 415/144 5845/1728 $\cdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\ddots$

For $i>0$, we have $a_n^i=\sum_{j=1}^{n}\frac{a_j^{i-1}}{j}=\sum_{j=0}^{i}\frac{a_{n-1}^{i-j}}{n^j}$, this can be concluded from $a_n^i=a_{n-1}^i+\frac{a_n^{i-1}}{n}$.

## Properties of $a_n^i$

From mathematical induction, we have $a_n^i\rightarrow\infty$ as $n\rightarrow\infty$. If $i\rightarrow\infty$, it is clear that

$a_n^i=\sum_{j=1}^n(-1)^{j-1}C_n^jj^{-i}=C_n^1-\frac{C_n^2}{2^i}+\cdots+\frac{(-1)^{n-1}C_n^n}{n^i}\rightarrow{n}.$

It is also clear that for fixed $i$ or $n$, $a_n^i$ monotonically increases.

For $i\ge{0}$, set the $(n-1)$th-order forward difference of $n^{-i},\cdots,2^{-i},1$ is $b_n^i$. By mathematical induction $b_n^i=\sum_{j=0}^{n-1}(-1)^jC_{n-1}^j(j+1)^{-i}$. Then

\begin{align*}
b_n^i&=\sum_{j=0}^{n-1}(-1)^jC_{n-1}^j(j+1)^{-i}\\
&=\sum_{j=1}^n(-1)^{j-1}C_{n-1}^{j-1}j^{-i}\\
&=\frac{1}{n}\sum_{j=1}^n(-1)^{j-1}C_n^jj^{-(i-1)}\\
&=\frac{1}{n}a_n^{i-1}.
\end{align*}

## Asymptotic expansion of $a_n^i$ as $n\rightarrow\infty$

For all $n\ge{1}$, $a_n^0=1$, so $a_n^1=\sum_{j=1}^n\frac{a_j^0}{j}=1+\frac{1}{2}+\cdots+\frac{1}{n}$, it is well known that when $n\rightarrow\infty$, $a_n^1\sim\ln{n}+\gamma$, where $\gamma$ is Euler-Mascheroni constant. Now naturally what is the asymptotic expansion of $a_n^i$ when $n\rightarrow\infty$?

From (1) we have

$a_n^i-a_{n-1}^i=\frac{a_n^{i-1}}{n}(n>1, i>0),$

regarding this equation as difference of $a_n^i$ for fixed $i$, when $n\rightarrow\infty$, if $a_n^{i-1}\sim{f_{i-1}(n)}$, we get

$f_i^{‘}(x)=\frac{1}{x}f_{i-1}(x).$

Because $a_n^0=1$, we set $f_0(x)=1$, so $f_1^{‘}(x)=\frac{1}{x}$, then $f_1(x)=\ln{x}+c$. We know $a_n^1=1+\frac{1}{2}+\cdots+\frac{1}{n}$, and $a_n^1\sim{\ln{n}+\gamma}$, so $f_1(x)$ is right, and $c=\gamma$. Now $f_2^{‘}(x)=\frac{1}{x}f_1(x)=\frac{\ln{x}+\gamma}{x}$, so $f_2(x)=\frac{1}{2}(\ln{x})^2+\gamma\ln{x}+c_1$. I have calculated $c_1$ on MATLAB by $a_n^2-\frac{1}{2}(\ln{n})^2-\gamma\ln{n}$, when $n=5000$, $c_1=0.98986542091089\cdots$. Now we let $\gamma_0=1$, $\gamma_1=\gamma$, $\gamma_2=c_1$, so $f_2(x)=\frac{\gamma_0}{2}(\ln{x})^2+\gamma_1\ln{x}+\gamma_2$. Continue doing this, we get $f_i(x)=\sum_{j=0}^i\frac{\gamma_{i-j}}{j!}(\ln{x})^j$, where $\gamma_k=\lim\limits_{n\rightarrow\infty}[a_n^k-\sum_{j=1}^k\frac{\gamma_{k-j}(\ln{n})^j}{j!}]$, and when $n\rightarrow\infty$, $a_n^i\sim{f_i(n)}$. Calculation on MATLAB gives $\gamma_3=0.90425195688241\cdots$, $\gamma_4=0.99113901006641\cdots$, $\gamma_5=0.95454545199027\cdots$, $\cdots$. We get a sequence $\{\gamma_k\}$ including Euler-Mascheroni constant $\gamma$.

Now we prove $a_n^i\sim{\sum_{j=0}^i\frac{\gamma_{i-j}}{j!}(\ln{n})^j}$, where $\gamma_k(0\le{k}\le{i})$ is defined before. Set $h_n^i=a_n^i-\sum_{j=1}^i\frac{\gamma_{i-j}}{j!}(\ln{n})^j$, we will prove the limit of $h_n^i$ exists.

$h_n^i\ge{0}$ and monotonically decreases as $n$ increases. So the limit of $h_n^i$ exists, set $\gamma_i=\lim\limits_{n\rightarrow\infty}h_n^i$ and

$u_n^i=h_n^i-\frac{1}{n}\sum_{j=0}^{i-1}\frac{\gamma_{i-1-j}(\ln{n})^j}{j!}, v_n^i=h_n^i-\frac{1}{4n}\sum_{j=0}^{i-1}\frac{\gamma_{i-1-j}(\ln{n})^j}{j!},$

for $i\ge{1}$ we have $u_n^i< \gamma_i< v_n^i$.

Proof. We prove the theorem by mathematical induction.

1) $h_n^0$ and $h_n^1$ satisfy the theorem.

Because $a_n^0=1$, and $\gamma_0=1$, so $h_n^0=0$ for all $n$, and is bounded.

Applying Lagrange mean value theorem on $\ln{x}$ between $[k-1,k]$ gives

$\ln{k}-\ln{(k-1)}=\frac{1}{k-1+\theta_k}<\frac{1}{k-1},$

where $0< \theta_k< 1$, sum from $k=2$ to $n$ we have $\ln{n}< \sum_{j=1}^{n-1}\frac{1}{j}$, that is $h_n^1=a_n^1-\ln{n}=\sum_{j=1}^n\frac{1}{j}-\ln{n}>\frac{1}{n}>0$, so $h_n^1$ is bounded below. For $n\ge{2}$ we have

\begin{align*}
h_n^1-h_{n-1}^1&=a_n^1-a_{n-1}^1+\ln{(n-1)}-\ln{n}\\
&=\frac{1}{n}a_n^0+\ln{(1-\frac{1}{n})}\\
&=\frac{1}{n}-(\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{3n^3}+\dots)\\
&=-(\frac{1}{2n^2}+\frac{1}{3n^3}+\dots)< 0 \end{align*}

so $h_n^1$ monotonically decreases. The limit of $h_n^1$ exists. Set $\lim\limits_{n\rightarrow\infty}h_n^1=\gamma_1$, we have $\lim\limits_{n\rightarrow\infty}u_n^1=\lim\limits_{n\rightarrow\infty}v_n^1=\gamma_1$, and

\begin{align*}
u_n^1-u_{n-1}^1&=\frac{1}{n}+\ln{(1-\frac{1}{n})}-\frac{1}{n}+\frac{1}{n-1}\\
&=\frac{1}{n-1}-(\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{3n^3}+\dots)\\
&=\frac{1}{n(n-1)}-\frac{1}{2n^2}-\frac{1}{3n^3}-\dots\\
&=\frac{1}{n}[\frac{1}{n-1}-\frac{1}{2n}-\frac{1}{3n^2}-\dots]\\
&=\frac{1}{n}[\frac{n+1}{2n(n-1)}-\frac{1}{3n^2}-\dots]\\
&=\frac{1}{n^2}[\frac{1}{2}+\frac{1}{n-1}-\frac{1}{3n}-\dots]>0,
\end{align*}

so if $n$ is large enough, $u_n^1-u_{n-1}^1>0$, $u_n^1$ monotonically increase to $\gamma_1$, i.e. $u_n^1< \gamma_1$. And

\begin{align*}
v_n^1-v_{n-1}&=\frac{1}{n}+\ln{(1-\frac{1}{n})}-\frac{1}{4n}+\frac{1}{4(n-1)}\\
&=-(\frac{1}{2n^2}+\frac{1}{3n^3}+\dots)+\frac{1}{4n(n-1)}\\
&=\frac{1}{n}[\frac{1}{4(n-1)}-\frac{1}{2n}-\frac{1}{3n^2}-\dots]\\
&=\frac{1}{n}[\frac{-n+2}{4n(n-1)}-\frac{1}{3n^2}-\dots]\\
&=\frac{1}{n^2}[-\frac{1}{4}+\frac{3}{4(n-1)}-\frac{1}{3n}-\dots]< 0, \end{align*}

so $v_n^1$ monotonically decreases to $\gamma_1$, i.e. $v_n^1>\gamma_1$. $h_n^1$ satisfies the theorem.

2) If $h_n^i(0\le{i}\le{k})$ satisfies the theorem, then $h_n^{k+1}$ satisfies the theorem

If for $i=k$, the proposition is right, i.e. $h_n^k\ge{0}$ and monotonically decreases as $n$ increases, and for $k\ge{1}$, we have

$u_n^k< \gamma_k< v_n^k.(2)$

Now we prove $h_n^{k+1}$ is bounded below and monotonically decreases as $n$ increases. Because $h_n^k\ge{0}$, we have $a_n^k\ge{\sum\limits_{j=1}^k\frac{\gamma_{k-j}}{j!}(\ln{n})^j}$. Applying Euler-Maclaurin Integration Formulas[3] gives

\begin{align*}
a_n^{k+1}&=\sum\limits_{j=1}^n\frac{a_j^k}{j}\le{\sum\limits_{j=1}^n\sum\limits_{m=0}^k\frac{\gamma_{k-m}}{m!}\frac{(\ln{j})^m}{j}}\\
&=\int_1^n{\sum\limits_{m=0}^k\frac{\gamma_{k-m}}{m!}\frac{(\ln{x})^m}{x}dx}+\frac{1}{2}(g(1)+g(n))\\
&+\sum\limits_{l=1}^{\infty}\frac{B_{2l}}{(2l)!}[g^{(2l-1)}(n)-g^{(2l-1)}(1)]\\
&=\sum\limits_{m=1}^{k+1}\frac{\gamma_{k+1-m}}{m!}(\ln{n})^m+\frac{1}{2}(\gamma_k+\sum\limits_{m=0}^k\frac{\gamma_{k-m}}{m!}\frac{(\ln{n})^m}{n})\\
&+\sum\limits_{l=1}^{\infty}\frac{B_{2l}}{(2l)!}[g^{(2l-1)}(n)-g^{(2l-1)}(1)]
\end{align*}

where $g(x)=\sum\limits_{m=0}^k\frac{\gamma_{k-m}}{m!}\frac{(\ln{x})^m}{x}$ and $B_{2l}$ is Bernoulli number, and we have

$g^{‘}(x)=\frac{1}{x^2}[\sum\limits_{m=0}^{k-1}\frac{\gamma_{k-1-m}(\ln{x})^m}{m!}-\sum\limits_{m=0}^k\frac{\gamma_{k-m}(\ln{x})^m}{m!}],$

so $g^{‘}(n)=-\frac{(\ln{n})^k}{n^2{k!}}+o(\frac{(\ln{n})^k}{n^2})$, and it is clear that $g^{(p)}(n)=o(\frac{(\ln{n})^k}{n^2})$ for $p>1$. For large $n$ we get

\begin{align*}
h_n^{k+1}&=a_n^{k+1}-\sum\limits_{m=1}^{k+1}\frac{\gamma_{k+1-m}(\ln{n})^m}{m!}\\
&\ge\frac{1}{2}[\gamma_k+\sum\limits_{m=0}^k\frac{\gamma_{k-m}}{m!}\frac{(\ln{n})^m}{n}]+\sum\limits_{l=1}^{\infty}\frac{B_{2l}}{(2l)!}[g^{(2l-1)}(n)-g^{(2l-1)}(1)]\\
&=\frac{1}{2}[\gamma_k+\sum\limits_{m=0}^k\frac{\gamma_{k-m}}{m!}\frac{(\ln{n})^m}{n}]+o(\frac{(\ln{n})^k}{n^2})>0
\end{align*}

so $h_n^{k+1}\ge{0}$.

For large $n$, applying equation (2) we have

\begin{align*}
h_n^{k+1}-h_{n-1}^{k+1}&=a_n^{k+1}-a_{n-1}^{k+1}+\sum\limits_{j=1}^{k+1}\frac{\gamma_{k+1-j}}{j!}\{[\ln{(n-1)}]^j-(\ln{n})^j\}\\
&\ge\frac{1}{n}a_n^k+\ln{(1-\frac{1}{n})}\sum\limits_{j=1}^{k+1}\frac{\gamma_{k+1-j}}{j!}\{[\ln{(n-1)}]^{j-1}+[\ln{(n-1)}]^{j-2}\ln{n}+\cdots+(\ln{n})^{j-1}\}\\
&< \frac{1}{n}[\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{n}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}]\\ &+\ln{(1-\frac{1}{n})}\sum\limits_{j=1}^{k+1}\frac{\gamma_{k+1-j}}{j!}\{[\ln{(n-1)}]^{j-1}+[\ln{(n-1)}]^{j-2}\ln{n}+\cdots+(\ln{n})^{j-1}\}\\ &=\frac{1}{n}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\ln{(1-\frac{1}{n})}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{n^2}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}\\ &+\ln{(1-\frac{1}{n})}\sum\limits_{j=1}^{k+1}\frac{\gamma_{k+1-j}}{j!}\{[\ln{(n-1)}]^{j-1}+[\ln{(n-1)}]^{j-2}\ln{n}+\cdots+(\ln{n})^{j-1}\}\\ &=-(\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots)\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{n^2}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}\\ &+\ln{(1-\frac{1}{n})}\sum\limits_{j=2}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=1}^{j-1}(\ln{n})^{j-1-q}\ln{(1-\frac{1}{n})}\sum\limits_{r=0}^{q-1}(\ln{n})^{q-1-r}[\ln{(n-1)}]^r\\ &=-(\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots)\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{n^2}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}\\ &+[\ln{(1-\frac{1}{n})}]^2\sum\limits_{j=2}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=1}^{j-1}(\ln{n})^{j-1-q}\sum\limits_{r=0}^{q-1}(\ln{n})^{q-1-r}[\ln{(n-1)}]^r\\ &=-\frac{(\ln{n})^k}{2n^2k!}+o(\frac{(\ln{n})^k}{n^2})+\frac{(\ln{n})^{k-1}}{n^2(k-1)!}+o(\frac{(\ln{n})^{k-1}}{n^2})+\frac{(\ln{n})^{k-1}}{n^2(k+1)!}+o(\frac{(\ln{n})^{k-1}}{n^2})\\ &=-\frac{(\ln{n})^k}{2n^2k!}+o(\frac{(\ln{n})^k}{n^2}), \end{align*}

so $h_n^{k+1}-h_{n-1}^{k+1}< 0$, i.e. $h_n^{k+1}$ monotonically decreases as n increases, and $h_n^{k+1}\ge{0}$, the limit of $h_n^{k+1}$ exists. Set $\lim\limits_{n\rightarrow\infty}h_n^{k+1}=\gamma_{k+1}$, we have $\lim\limits_{n\rightarrow\infty}u_n^{k+1}=\lim\limits_{n\rightarrow\infty}v_n^{k+1}=\gamma_{k+1}$, and

\begin{align*}
u_n^{k+1}-u_{n-1}^{k+1}&=\frac{1}{n}a_n^k+\ln{(1-\frac{1}{n})}\sum\limits_{j=1}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=0}^{j-1}(\ln{n})^{j-1-q}[\ln{(n-1)}]^q\\
&-\frac{1}{n}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{n-1}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}[\ln{(n-1)}]^j}{j!}\\
&>\frac{1}{n}[\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{4n}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}]\\
&+\ln{(1-\frac{1}{n})}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}\\
&+\ln{(1-\frac{1}{n})}\sum\limits_{j=2}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=1}^{j-1}(\ln{n})^{j-1-q}\ln{(1-\frac{1}{n})}\sum\limits_{r=0}^{q-1}(\ln{n})^r[\ln{(n-1)}]^{q-1-r}\\
&+\frac{1}{n(n-1)}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}\\
&+\frac{\ln{(1-\frac{1}{n})}}{n-1}\sum\limits_{j=1}^{k}\frac{\gamma_{k-j}}{j!}\sum\limits_{q=0}^{j-1}(\ln{n})^{j-1-q}[\ln{(n-1)}]^q\\
&=-(\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots)\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{4n^2}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}\\
&+[\ln{(1-\frac{1}{n})}]^2\sum\limits_{j=2}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=1}^{j-1}(\ln{n})^{j-1-q}\sum\limits_{r=0}^{q-1}(\ln{n})^{r}[\ln{(n-1)}]^{q-1-r}\\
&+\frac{1}{n(n-1)}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}\\
&+\frac{\ln{(1-\frac{1}{n})}}{n-1}\sum\limits_{j=1}^{k}\frac{\gamma_{k-j}}{j!}\sum\limits_{q=0}^{j-1}(\ln{n})^{j-1-q}[\ln{(n-1)}]^q\\
&=\frac{1}{n(n-1)k!}(\ln{n})^k-\frac{1}{2n^2k!}(\ln{n})^k+o(\frac{(\ln{n})^k}{n^2})\\
&=\frac{(n+1)(\ln{n})^k}{2n^2(n-1)k!}+o(\frac{(\ln{n})^k}{n^2})\\
&=\frac{(\ln{n})^k}{2n^2k!}+o(\frac{(\ln{n})^k}{n^2})>0,
\end{align*}

so for large $n$, $u_n^{k+1}$ monotonically increases to $\gamma_{k+1}$, i.e. $u_n^{k+1}< \gamma_{k+1}$. And

\begin{align*}
v_n^{k+1}-v_{n-1}^{k+1}&=\frac{1}{n}a_n^k+\ln{(1-\frac{1}{n})}\sum\limits_{j=1}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=0}^{j-1}(\ln{n})^{j-1-q}[\ln{(n-1)}]^q\\
&-\frac{1}{4n}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{4(n-1)}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}[\ln{(n-1)}]^j}{j!}\\
&< \frac{1}{n}[\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{n}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}]\\ &+\ln{(1-\frac{1}{n})}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}\\ &+\ln{(1-\frac{1}{n})}\sum\limits_{j=2}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=1}^{j-1}(\ln{n})^{j-1-q}\ln{(1-\frac{1}{n})}\sum\limits_{r=0}^{q-1}(\ln{n})^r[\ln{(n-1)}]^{q-1-r}\\ &+\frac{1}{4n(n-1)}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}\\ &+\frac{\ln{(1-\frac{1}{n})}}{4(n-1)}\sum\limits_{j=1}^{k}\frac{\gamma_{k-j}}{j!}\sum\limits_{q=0}^{j-1}(\ln{n})^{j-1-q}[\ln{(n-1)}]^q\\ &=-(\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots)\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}+\frac{1}{4n^2}\sum\limits_{j=0}^{k-1}\frac{\gamma_{k-1-j}(\ln{n})^j}{j!}\\ &+[\ln{(1-\frac{1}{n})}]^2\sum\limits_{j=2}^{k+1}\frac{\gamma_{k+1-j}}{j!}\sum\limits_{q=1}^{j-1}(\ln{n})^{j-1-q}\sum\limits_{r=0}^{q-1}(\ln{n})^{r}[\ln{(n-1)}]^{q-1-r}\\ &+\frac{1}{4n(n-1)}\sum\limits_{j=0}^{k}\frac{\gamma_{k-j}(\ln{n})^j}{j!}\\ &+\frac{\ln{(1-\frac{1}{n})}}{4(n-1)}\sum\limits_{j=1}^{k}\frac{\gamma_{k-j}}{j!}\sum\limits_{q=0}^{j-1}(\ln{n})^{j-1-q}[\ln{(n-1)}]^q\\ &=\frac{1}{4n(n-1)k!}(\ln{n})^k-\frac{1}{2n^2k!}(\ln{n})^k+o(\frac{(\ln{n})^k}{n^2})\\ &=\frac{(-n+2)(\ln{n})^k}{4n^2(n-1)k!}+o(\frac{(\ln{n})^k}{n^2})\\ &=-\frac{(\ln{n})^k}{4n^2k!}+o(\frac{(\ln{n})^k}{n^2})< 0, \end{align*}

so for large $n$, $v_n^{k+1}$ monotonically decreases to $\gamma_{k+1}$, i.e. $v_n^{k+1}>\gamma_{k+1}$. $h_n^{k+1}$ satisfies the theorem.

From 1) and 2), we proved Theorem 1. So $h_n^i$ converges. Set

$\gamma_i=\lim\limits_{n\rightarrow\infty}h_n^i=\lim\limits_{n\rightarrow\infty}[a_n^i-\sum\limits_{j=1}^i\frac{\gamma_{i-j}(\ln{n})^j}{j!}],$

we have $a_n^i\sim{\sum\limits_{j=0}^i\frac{\gamma_{i-j}}{j!}(\ln{n})^j}$ as $n\rightarrow\infty$.

Because $b_n^{i+1}=\frac{a_n^i}{n}$, so

$\lim_{n\rightarrow\infty}b_n^{i+1}=\lim_{n\rightarrow\infty}\sum_{j=0}^i\frac{\gamma_{i-j}}{j!}\frac{(\ln{n})^j}{n}=0.$

We have $I_{m,n}(x)=\frac{m!x^n}{n!}\sum_{i=0}^m\frac{a_n^i(-\ln{x})^{m-i}}{(m-i)!}$ defined before, when $n\rightarrow\infty$, $a_n^i\sim{\sum_{j=0}^i\frac{\gamma_{i-j}}{j!}(\ln{n})^j}$, so

$I_{m,n}(n)=\frac{m!n^n}{n!}\sum_{i=0}^m\frac{a_n^i(-\ln{n})^{m-i}}{(m-i)!}\sim\frac{m!n^n}{n!}\sum_{i=0}^m\sum_{j=0}^i\frac{(-1)^{m-i}\gamma_{i-j}(\ln{n})^{m+j-i}}{j!(m-i)!},$

if $m+j-i=0$, $i=m+j$, and $i\le{m}$, so $j=0$, the coefficient of $(\ln{n})^0$ is $\gamma_m$. If $m+j-i=k>0$, then $i-j=m-k$, we have $\gamma_{i-j}=\gamma_{m-k}$. If $i< m-k$, $m+j-i>k+j\ge{k}$, so when $i$ increases from $m-k$ to $m$, $j$ increases from $0$ to $k$. Summate the terms with $(\ln{n})^k$ and for $1\le{k}\le{m}$ get

\begin{align*}
\sum_{j=0}^k\frac{(-1)^{k-j}\gamma_{m-k}(\ln{n})^k}{j!(k-j)!}&=\frac{\gamma_{m-k}(\ln{n})^k}{k!}\sum_{j=0}^k\frac{(-1)^{k-j}k!}{j!(k-j)!}\\
&=\frac{\gamma_{m-k}(\ln{n})^k}{k!}\sum_{j=0}^k(-1)^{k-j}C_k^j\\
&=0,
\end{align*}

so $I_{m,n}(n)\sim{\frac{m!n^n}{n!}\gamma_m}$, $n\rightarrow\infty$, that is $\gamma_m=\lim\limits_{n\rightarrow\infty}\sum\limits_{i=0}^m\frac{a_n^i(-\ln{n})^{m-i}}{(m-i)!}$. I have calculated $\gamma_m$ using this formula in JAVA program and get some values of $\gamma_m$ below:

\begin{align*}
&\gamma_0=1.0,\\
&\gamma_1=0.5772157148989514\cdots,\\
&\gamma_2=0.9890559742083838\cdots,\\
&\gamma_3=0.9074790967084709\cdots,\\
&\gamma_4=0.9817280811976161\cdots,\\
&\gamma_5=0.981995086904135\cdots,\\
&\gamma_6=0.9931490272210795\cdots,\\
&\gamma_7=0.9960021871229401\cdots,\\
&\gamma_8=0.998103952704696\cdots,\\
&\gamma_9=0.999031200830359\cdots,\\
&\gamma_{10}=0.9994992676656693\cdots,
\end{align*}

I guess for $m>0$, $0<\gamma_m<1$ and $\gamma_m$ is irrational.

Generalizing $a_n^i$, set $a_n^s=\sum_{j=1}^n(-1)^{j-1}C_n^jj^{-s}$, $s\ge{0}$. Then we still have the recurrence equation $a_n^s=a_{n-1}^s+\frac{1}{n}a_n^{s-1}$, now we prove it.

\begin{align*}
a_n^s-a_{n-1}^s&=\sum_{j=1}^n(-1)^{j-1}C_n^jj^{-s}-\sum_{j=1}^{n-1}(-1)^{j-1}C_{n-1}^jj^{-s}\\
&=\sum_{j=1}^{n-1}(-1)^{j-1}[C_n^j-C_{n-1}^j]j^{-s}+(-1)^{n-1}C_n^nn^{-s}\\
&=\sum_{j=1}^{n-1}(-1)^{j-1}[\frac{n!}{j!(n-j)!}-\frac{(n-1)!}{j!(n-1-j)!}]j^{-s}+(-1)^{n-1}C_n^nn^{-s}\\
&=\sum_{j=1}^{n-1}(-1)^{j-1}[\frac{n}{n-j}-1]\frac{(n-1)!}{j!(n-1-j)!}j^{-s}+(-1)^{n-1}C_n^nn^{-s}\\
&=\sum_{j=1}^{n-1}(-1)^{j-1}\frac{(n-1)!j}{j!(n-j)!}j^{-s}+(-1)^{n-1}C_n^nn^{-s}\\
&=\frac{1}{n}\sum_{j=1}^{n-1}(-1)^{j-1}\frac{n!}{j!(n-j)!}j^{-(s-1)}+(-1)^{n-1}\frac{1}{n}C_n^nn^{-(s-1)}\\
&=\frac{1}{n}\sum_{j=1}^{n}(-1)^{j-1}\frac{n!}{j!(n-j)!}j^{-(s-1)}\\
&=\frac{1}{n}a_n^{s-1}.
\end{align*}

So $a_n^s=a_{n-1}^s+\frac{1}{n}a_n^{s-1}$. Because $a_1^s=1$, obviously $a_n^s=\sum_{j=1}^n\frac{a_j^{s-1}}{j}$ still holds.

## References

[1] 华东师范大学数学系编. 数学分析[上册]. 北京: 高等教育出版社, 2001, 187.

[2] 孙淑玲, 许胤龙. 组合数学引论. 合肥: 中国科学技术大学出版社, 1999, 176.

[3] http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html.