The Log-Gamma Distribution and Non-Normal Error

Appendices

Appendix A. The gamma, log-gamma, and polygamma functions

A.1. The gamma function as an integral

The modern form of the gamma function is Γ(α) = $\int_{x=0}^{\infty}$ e^−xx^α−1dx. The change of variable t = e^−x (or x = −ln t = ln 1/t) transforms it into:

$$\Gamma(\alpha)=\int_{t=1}^{0} t\left(\ln \frac{1}{t}\right)^{\alpha-1}\left(-\frac{d t}{t}\right)=\int_{t=0}^{1}\left(\ln \frac{1}{t}\right)^{\alpha-1} d t$$

According to Havil [2003, p. 53], Leonhard Euler used the latter form in his pioneering work during 1729–1730. It was not until 1809 that Legendre named it the gamma function and denoted it with the letter ‘Γ’. The function records the struggle between the exponential function e^−x and the power function x^α−1, in which the former ultimately prevails and forces the convergence of the integral for α > 0.

Of course, $\Gamma(1)=\int_{x=0}^{\infty} e^{-x} d x=1$. The well-known recurrence formula Γ(α + 1) = αΓ(α) follows from integration by parts:

$$\begin{aligned} \alpha \Gamma(\alpha) & =\alpha \int_{x=0}^{\infty} e^{-x} x^{\alpha-1} d x=\int_{x=0}^{\infty} e^{-x} d\left(x^{\alpha}\right)=\left.e^{-x} x^{\alpha}\right|_{0} ^{\infty} \\ & -\int_{x=0}^{\infty} x^{\alpha} d\left(e^{-x}\right)=0+\int_{x=0}^{\infty} e^{-x} x^{\alpha+1-1} d x=\Gamma(\alpha+1) \end{aligned}$$

It is well to note that in order for e^−xx^α|₀^∞ to equal zero, α must be positive. For positive-integral values of α, Γ(α) = (α − 1)!; so the gamma function extends the factorial to the positive real numbers. In this domain the function is continuous, even differentiable, as well as positive. Although $\lim _{\alpha \rightarrow 0^{+}} \Gamma(\alpha)=\infty$, $\lim _{\alpha \rightarrow 0^{+}} \alpha \Gamma(\alpha)=\lim _{\alpha \rightarrow 0} \Gamma(\alpha+1)=\Gamma(1)=1$. So $\Gamma(\alpha) \sim \frac{1}{\alpha}$ as α → 0⁺. As α increases away from zero, the function decreases to a minimum of approximately Γ(1.462) = 0.886, beyond which it increases. So, over the positive real numbers the gamma function is ‘U’ shaped, or concave upward.

The simple algebra $1=\frac{\Gamma(\alpha)}{\Gamma(\alpha)}=\frac{\int_{x=0}^{\infty} e^{-x} x^{\alpha-1} d x}{\Gamma(\alpha)}=$ $\int_{x=0}^{\infty} \frac{1}{\Gamma(\alpha)} e^{-x} x^{\alpha-1} d x$ indicates that $f(x)=\frac{1}{\Gamma(\alpha)} e^{-x} x^{\alpha-1}$ is the probability density function of what we will call a Gamma(α, 1)-distributed random variable, or simply Gamma(α)-distributed with θ = 1 understood.^[16] For k > −α, the kth moment of X ∼ Gamma(α) is:

$$\begin{aligned} E\left[X^{k}\right] & =\int_{x=0}^{\infty} x^{k} \frac{1}{\Gamma(\alpha)} e^{-x} x^{\alpha-1} d x \\ & =\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)} \int_{x=0}^{\infty} \frac{1}{\Gamma(\alpha+k)} e^{-x} x^{\alpha+k-1} d x \\ & =\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)} \end{aligned}$$

Therefore, for k = 1 and 2, E[X] = α and E[X²] = α(α + 1). Hence, Var[X] = α.

A.2. The gamma function as an infinite product

Euler found that the gamma function could be expressed as the infinite product:

$$\Gamma(x)=\lim _{n \rightarrow \infty} \frac{n!n^{x}}{x(1+x) \ldots(n+x)}$$

$$=\lim _{n \rightarrow \infty} \frac{n^{x}}{x\left(1+\frac{x}{1}\right) \ldots\left(1+\frac{x}{n}\right)}$$

As a check, $\Gamma(1)=\lim _{n \rightarrow \infty} \frac{n!n}{(n+1)!}=\lim _{n \rightarrow \infty} \frac{n}{n+1}=1$. Moreover, the recurrence formula is satisfied:

$$\begin{aligned} \Gamma(x+1)= & \lim _{n \rightarrow \infty} \frac{n!n^{x+1}}{(1+x) \ldots(n+x)(n+1+x)} \\ = & x \lim _{n \rightarrow \infty} \frac{n!n^{x}}{x(1+x) \ldots(n+x)} \\ & \lim _{n \rightarrow \infty} \frac{n}{(n+1+x)}=x \Gamma(x) \end{aligned}$$

Hence, by induction, the integral and infinite-product definitions are equivalent for positive-integral values of x. It is the purpose of this section to extend the equivalence to all positive real numbers.

The proof is easier in logarithms. The log-gamma function is $\ln \Gamma(\alpha)=\ln \int_{x=0}^{\infty} e^{-x} x^{\alpha-1} d x$. The form of its recursive formula is lnΓ(α + 1) = ln(αΓ(α)) = lnΓ(α) + ln α. Its first derivative is:

$$\begin{aligned} (\ln \circ \Gamma)^{\prime}(\alpha) & =\frac{d}{d \alpha} \ln \Gamma(\alpha) \equiv \frac{\Gamma^{\prime}(\alpha)}{\Gamma(\alpha)} \\ & =\frac{\int_{x=0}^{\infty} e^{-x} x^{\alpha-1} \ln x d x}{\Gamma(\alpha)} \\ & =\int_{x=0}^{\infty} \frac{1}{\Gamma(\alpha)} e^{-x} x^{\alpha-1} \ln x d x=E[Y] \end{aligned}$$

So the first derivative of lnΓ(α) is the expectation of Y ∼ Log-Gamma(α), as defined in Section 2. The second derivative is:

$$\begin{aligned} (\ln \circ \Gamma)^{\prime \prime}(\alpha) & =\frac{d}{d \alpha} \frac{\Gamma^{\prime}(\alpha)}{\Gamma(\alpha)} \\ & =\frac{\Gamma(\alpha) \Gamma^{\prime \prime}(\alpha)-\Gamma^{\prime}(\alpha) \Gamma^{\prime}(\alpha)}{\Gamma(\alpha) \Gamma(\alpha)} \end{aligned}$$

$$\begin{array}{l} =\frac{\Gamma^{\prime \prime}(\alpha)}{\Gamma(\alpha)}-\left(\frac{\Gamma^{\prime}(\alpha)}{\Gamma(\alpha)}\right)^{2} \\ =\int_{x=0}^{\infty} \frac{1}{\Gamma(\alpha)} e^{-x} x^{\alpha-1}(\ln x)^{2} d x-E[Y]^{2} \\ =E\left[Y^{2}\right]-E[Y]^{2}=\operatorname{Var}[Y]>0 \end{array}$$

Because the second derivative is positive, the log-gamma function is concave upward over the positive real numbers.

Now let n ∈ {2, 3, . . .} and x ∈ (0, 1]. So 0 < n − 1 < n < n + x ≤ n + 1. We consider the slope of the log-gamma secants over the intervals [n – 1, n], [n, n + x], and [n, n + 1]. Due to the upward concavity, the order of these slopes must be:

$$\begin{array}{c} \frac{\ln \Gamma(n)-\ln \Gamma(n-1)}{n-(n-1)}<\frac{\ln \Gamma(n+x)-\ln \Gamma(n)}{(n+x)-n} \\ \leq \frac{\ln \Gamma(n+1)-\ln \Gamma(n)}{(n+1)-n} \end{array}$$

Using the recurrence formula and multiplying by positive x, we continue:

$$\begin{array}{c} x \ln (n-1)<\ln \Gamma(n+x)-\ln \Gamma(n) \leq x \ln (n) \\ x \ln (n-1)+\ln \Gamma(n)<\ln \Gamma(n+x) \leq x \ln (n)+\ln \Gamma(n) \end{array}$$

But lnΓ(n) = $\sum_{k=1}^{n-1} \ln k$ and ln Γ(n + x) = lnΓ(x) + ln x +

$\sum_{k=1}^{n-1} \ln (k+x)$. Hence:

$$\begin{aligned} \begin{aligned} x & \ln (n-1)+\sum_{k=1}^{n-1} \ln k \\ & < \end{aligned} & \ln \Gamma(x)+\ln x \\ & \quad+\sum_{k=1}^{n-1} \ln (k+x) \leq x \ln (n)+\sum_{k=1}^{n-1} \ln k \\ x \ln (n-1)+\sum_{k=1}^{n-1} \ln k & -\ln x-\sum_{k=1}^{n-1} \ln (k+x)<\ln \Gamma(x) \\ & \leq x \ln (n)+\sum_{k=1}^{n-1} \ln k-\ln x-\sum_{k=1}^{n-1} \ln (k+x) \end{aligned}$$

$$\begin{aligned} x \ln (n-1)-\ln x-\sum_{k=1}^{n-1} & \ln \left(1+\frac{x}{k}\right)<\ln \Gamma(x) \\ & \leq x \ln (n)-\ln x-\sum_{k=1}^{n-1} \ln \left(1+\frac{x}{k}\right) \end{aligned}$$

This string of inequalities, in the middle of which is ln Γ(x), is true for n ∈{2, 3, . . .}. As n approaches infinity, $\lim _{n \rightarrow \infty}$[x ln(n) – x ln(n − 1)] $=x \lim _{n \rightarrow \infty} \ln \left(\frac{n}{n-1}\right)$ = $x \ln \left(\lim _{n \rightarrow \infty} \frac{n}{n-1}\right)$ = xln1 = 0. So lnΓ(x) is sandwiched between two series that have the same limit. Hence:

$$\begin{aligned} \ln \Gamma(x) & =-\ln x+\lim _{n \rightarrow \infty}\left\{x \ln (n-1)-\sum_{k=1}^{n-1} \ln \left(1+\frac{x}{k}\right)\right\} \\ & =-\ln x+\lim _{n \rightarrow \infty}\left\{x \ln (n)-\sum_{k=1}^{n} \ln \left(1+\frac{x}{k}\right)\right\} \end{aligned}$$

If the limit did not converge, then lnΓ(x) would not converge. Nevertheless, Appendix C constructs a proof of the limit’s convergence. But for now we can finish this section by exponentiation:

$$\begin{aligned} \Gamma(x) & =e^{\ln \Gamma(x)} \\ & =\frac{1}{x} e^{\lim _{n \rightarrow \infty}\left\{x \ln (n)-\sum_{k=1}^{n} \ln \left(1+\frac{x}{k}\right)\right\}} \\ & =\frac{1}{x} \lim _{n \rightarrow \infty}\left\{e^{x \ln (n)-\sum_{k=1}^{n} \ln \left(1+\frac{x}{k}\right)}\right\} \\ & =\frac{1}{x} \lim _{n \rightarrow \infty} \frac{n^{x}}{\prod_{k=1}^{n}\left(1+\frac{x}{k}\right)} \\ & =\lim _{n \rightarrow \infty} \frac{n^{x}}{x\left(1+\frac{x}{1}\right) \ldots\left(1+\frac{x}{n}\right)} \end{aligned}$$

The infinite-product form converges for all complex numbers z ∉ {0, −1, −2, . . .}; it provides the analytic continuation to the integral form. ^[17] With the infinite-product form, Havil [2003, pp. 58f] easily derives the complement formula Γ(x)Γ(1 − x) = π/sin(πx). Consequently, Γ(½)Γ(½) = π/sin(π/2) = π.

A.3. Derivatives of the log-gamma function

The derivative of the gamma function is Γ′(a) = $\frac{d}{d \alpha} \int_{x=0}^{\infty} e^{-x} x^{\alpha-1} d x=\int_{x=0}^{\infty} e^{-x} \frac{d\left(x^{\alpha-1}\right)}{d \alpha} d x=\int_{x=0}^{\infty} e^{-x} x^{\alpha-1} \ln x d x$

So, for X ∼ Gamma(α):

$$\begin{aligned} E[\ln X] & =\int_{x=0}^{\infty} \ln x \frac{1}{\Gamma(\alpha)} e^{-x} x^{\alpha-1} d x \\ & =\frac{1}{\Gamma(\alpha)} \int_{x=0}^{\infty} e^{-x} x^{\alpha-1} \ln x d x \\ & =\frac{\Gamma^{\prime}(\alpha)}{\Gamma(\alpha)}=\frac{d}{d \alpha} \ln \Gamma(\alpha) \end{aligned}$$

Because the derivative of ln Γ(α) has proven useful, mathematicians have named it the “digamma” function: $\psi(\alpha)=\frac{d}{d \alpha} \ln \Gamma(\alpha)$. Therefore, E[ln X] = ψ(α). Its derivatives are the trigamma, tetragamma, and so on. Unfortunately, Excel does not provide these functions, although one can approximate them from GAMMALN by numerical differentiation. The R and SAS programming languages have the trigamma function but stop at the second derivative of ln Γ(α). The remainder of this appendix deals with the derivatives of the log-gamma function.

According to the previous section:

$$\ln \Gamma(x)=-\ln x+\lim _{n \rightarrow \infty}\left\{x \ln (n)-\sum_{k=1}^{n} \ln \left(1+\frac{x}{k}\right)\right\}$$

Hence the digamma function is:

$$\begin{aligned} \psi(x) & =\frac{d}{d x} \ln \Gamma(x)=-\frac{1}{x}+\lim _{n \rightarrow \infty}\left\{\ln n-\sum_{k=1}^{n} \frac{1}{k+x}\right\} \\ & =\lim _{n \rightarrow \infty}\left\{\ln n-\sum_{k=0}^{n} \frac{1}{k+x}\right\} \end{aligned}$$

Its recurrence formula is:

$$\begin{aligned} \psi(x+1) & =\lim _{n \rightarrow \infty}\left\{\ln n-\sum_{k=0}^{n} \frac{1}{k+1+x}\right\} \\ & =\lim _{n \rightarrow \infty}\left\{\ln n-\sum_{k=0}^{n} \frac{1}{k+x}\right\}+\frac{1}{0+x}=\psi(x)+\frac{1}{x} \end{aligned}$$

Also, $\psi(1)=\lim _{n \rightarrow \infty}\left\{\ln n-\sum_{k=0}^{n} \frac{1}{k+1}\right\}=\lim _{n \rightarrow \infty}$ $\left\{\ln n-\sum_{k=1}^{n+1} \frac{1}{k}\right\}=-\lim _{n \rightarrow \infty}\left\{\sum_{k=1}^{n} \frac{1}{k}-\ln n\right\}=-\gamma$. Euler was

the first to see that $\lim _{n \rightarrow \infty}\left\{\sum_{k=1}^{n} \frac{1}{k}-\ln n\right\}$ converged.^[18] He named it ‘C’ and estimated its value as 0.577218. Eventually it was called the Euler-Mascheroni constant γ, whose value to six places is really 0.577216.

The successive derivatives of the digamma function are:

$$\begin{aligned} \psi^{\prime}(x) & =\frac{d}{d x} \lim _{n \rightarrow \infty}\left\{\ln n-\sum_{k=0}^{n} \frac{1}{k+x}\right\}=\sum_{k=0}^{\infty} \frac{1}{(k+x)^{2}} \\ \psi^{\prime \prime}(x) & =-2 \sum_{k=0}^{\infty} \frac{1}{(k+x)^{3}} \\ \psi^{\prime \prime \prime}(x) & =6 \sum_{k=0}^{\infty} \frac{1}{(k+x)^{4}} \end{aligned}$$

The general formula for the nth derivative is ψ^[n](x) = (−1)ⁿ⁻¹n!$\sum_{k=0}^{\infty} \frac{1}{(k+x)^{n+1}}$ with recurrence formula:

$$\begin{aligned} \psi^{[n]}(x+1) & =(-1)^{n-1} n!\sum_{k=0}^{\infty} \frac{1}{(k+1+x)^{n+1}} \\ & =(-1)^{n-1} n!\sum_{k=0}^{\infty} \frac{1}{(k+x)^{n+1}}-\frac{(-1)^{n-1} n!}{x^{n+1}} \\ & =\psi^{[n]}(x)+\frac{(-1)^{n} n!}{x^{n+1}} \end{aligned}$$

Finally, ψ^[n](1) = (−1)ⁿ⁻¹n!$\sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}}=(-1)^{n-1}$ $n!\sum_{k=1}^{\infty} \frac{1}{k^{n+1}}=(-1)^{n-1} n!\zeta(n+1)$. A graph of the digamma through pentagamma functions appears in Wikipedia [Polygamma function].

Appendix B. Moments versus cumulants^[19]

Actuaries and statisticians are well aware of the moment generating function (mgf) of random variable $X$:

$$M_X(t)=E\left[e^{t X}\right]=\int_X e^{t x} f(x) d x$$

Obviously, $M_X(0)=E[1]=1$. It is called the mgf because its derivatives at zero, if convergent, equal the (non-central) moments of $X$, i.e.:

$$\begin{aligned} M_X^{[n]}(0) & =\left.\frac{d^n}{d t^n} E\left[e^{t X}\right]\right|_{t=0}=E\left[\left.\frac{d^n e^{t X}}{d t^n}\right|_{t=0}\right] \\ & =E\left[X^n e^{0 \cdot X}\right]=E\left[X^n\right] . \end{aligned}$$

The mgf for the normal random variable is $M_{N\left(\mu, \sigma^2\right)}(t)=e^{\mu t+\sigma^2 t^2 / 2}$. The mgf of a sum of independent random variables equals the product of their mgfs:

$$\begin{aligned} M_{X+Y}(t) & =E\left[e^{t(X+Y)}\right]=E\left[e^{t X} e^{t Y}\right] \\ & =E\left[e^{t X}\right] E\left[e^{t Y}\right]=M_X(t) M_Y(t) \end{aligned}$$

The cumulant generating function (cgf) of $X$, or $\psi_X(t)$ is the natural logarithm of the moment generating function. So $\psi_X(0)=0$; and for independent summands:

$$\begin{aligned} \psi_{X+Y}(t) & =\ln M_{X+Y}(t)=\ln M_X(t) M_Y(t) \\ & =\ln M_X(t)+\ln M_Y(t)=\psi_X(t)+\psi_Y(t) \end{aligned}$$

The $n$th cumulant of $X$, or $\kappa_n(X)$, is the $n$th derivative of the cgf evaluated at zero. The first two cumulants are identical to the mean and the variance:

$$\begin{aligned} \kappa_1(X)= & \left.\frac{d \ln M_X(t)}{d t}\right|_{t=0}=\left.\frac{M_X^{\prime}(t)}{M_X(t)}\right|_{t=0}=E[X] \\ \kappa_2(X) & =\left.\frac{d}{d t} \frac{M_X^{\prime}(t)}{M_X(t)}\right|_{t=0}=\left.\frac{-M_X^{\prime}(t) M_X^{\prime \prime}(t)}{M_X^2(t)}\right|_{t=0} ^{\prime}(t) \\ = & E\left[X^2\right]-E[X] E[X]=\operatorname{Var}[X] \end{aligned}$$

One who performs the somewhat tedious third derivative will find that the third cumulant is identical to the skewness. So the first three cumulants are equal to the first three central moments. This amounts to a proof that the first three central moments are additive.

However, the fourth and higher cumulants do not equal their corresponding central moments. In fact, defining the central moments as $\mu_n=E\left[(X-\mu)^n\right]$, the next three cumulant relations are:

$$\begin{aligned} & \kappa_4=\mu_4-3 \mu_2^2=\mu_4-3 \sigma^4 \\ & \kappa_5=\mu_5-10 \mu_2 \mu_3 \\ & \kappa_6=\mu_6-15 \mu_2 \mu_4-10 \mu_3^2+30 \mu_2^3 \end{aligned}$$

The cgf of a normal random variable is $\psi_{N\left(\mu, \sigma^2\right)}(t)= \ln e^{\mu t+\sigma^2 t / 2}=\mu t+\sigma^2 t^2 / 2$. It first two cumulants equal $\mu$ and $\sigma^2$, and its higher cumulants are zero. Therefore, the third and higher cumulants are relative to the zero values of the corresponding normal cumulants. So third and higher cumulants could be called “excess of the normal,” although this in practice is done only for the fourth cumulant. Because $\mu_4=E\left[(X-\mu)^4\right]$ is often called the kurtosis, ambiguity is resolved by calling $\kappa_4$ the “excess kurtosis,” or the kurtosis in excess of $\mu_4=E\left[N\left(0, \sigma^2\right)^4\right]=E\left[N(0,1)^4\right] \cdot \sigma^4=3 \sigma^4$.

Appendix C. Log-gamma convergence

In Appendix A. 2 we proved that for real $\alpha>0$ :

$$\begin{aligned} \ln \left(\int_{x=0}^{\infty} e^{-x} x^{\alpha-1} d x\right)= & \ln \Gamma(\alpha)=-\ln \alpha \\ & +\lim _{n \rightarrow \infty}\left\{\alpha \ln n-\sum_{k=1}^n \ln \left(1+\frac{\alpha}{k}\right)\right\} \end{aligned}$$

There we argued that the “real-ness” of the left side for all $\alpha>0$ implies the convergence of the limit on the right side; otherwise, the equality would be limited or qualified. Nevertheless, it is valuable to examine the convergence of the log-gamma limit, as we will now do.

Define $a_n(x)=\frac{x}{n}-\ln \left(1+\frac{x}{n}\right)$ for natural number $n$ and real number $x$. To avoid logarithms of negative numbers, $x>-1$, or $x \in(-1, \infty)$. The root of this group of functions is $f(x)=x-\ln (1+x)$. Since the two functions $y=x$ and $y=\ln (1+x)$ are tangent to each other at $x=0$, $f(0)=f^{\prime}(0)=0$. Otherwise, the line is above the logarithm and $f(x \neq 0)>0$. Since $f^{\prime \prime}(x)=1 /(1+x)^2$ must be positive, $f(x)$ is concave upward with a global minimum over $(-1, \infty)$ at $x=0$.

For now, let us investigate the convergence of $\varphi(x)=\sum_{n=1}^{\infty} a_n(x)=\sum_{n=1}^{\infty}\left\{\frac{x}{n}-\ln \left(1+\frac{x}{n}\right)\right\}$. Obviously, $\varphi(x=0)=0$. But if $x \neq 0$, for some $\xi \in(\min (n, n+x)$, $\max (n, n+x)$ ):

$$\begin{aligned} a_n(x) & =\frac{x}{n}-\ln \left(1+\frac{x}{n}\right) \\ & =\frac{x}{n}-\ln \left(\frac{n+x}{n}\right) \\ & =\frac{x}{n}-\int_{u=n}^{n+x} \frac{d u}{u} \\ & =\frac{x}{n}-((n+x)-n) \frac{1}{\xi} \\ & =\frac{x}{n}-\frac{x}{\xi} \\ & =x\left(\frac{1}{n}-\frac{1}{\xi}\right) \end{aligned}$$

Regardless of the sign of $x$ :

$$|x|\left|\left(\frac{1}{n}-\frac{1}{n}\right)\right|<\left|a_n(x)\right|<|x|\left|\left(\frac{1}{n}-\frac{1}{n+x}\right)\right|$$

The lower bound reduces to zero; the upper to $|x|\left|\frac{x}{n(n+x)}\right|=\frac{|x|^2}{|n||n+x|}=\frac{x^2}{n|n+x|}$. But $n+x$ must be positive, since $x>-1$. Therefore, for $x \neq 0$ :

$$0<\left|a_n(x)\right|<\frac{x^2}{n(n+x)}$$

For $x=0$, equality prevails. And so for all $x>-1$ :

$$0 \leq\left|a_n(x)\right| \leq \frac{x^2}{n(n+x)}$$

Now absolute convergence, or the convergence of $\sum_{n=1}^{\infty}\left|a_n(x)\right|$, is stricter than simple convergence. If $\sum_{n=1}^{\infty}\left|a_n(x)\right|$ converges, then so too does $\varphi(x)=$ $\sum_{n=1}^{\infty} a_n(x)$. And $\sum_{n=1}^{\infty}\left|a_n(x)\right|$ converges, if it is bounded from above. The following string of inequalities establishes the upper bound:

$$\begin{aligned} \sum_{n=1}^{\infty}\left|a_n(x)\right| & \leq \sum_{n=1}^{\infty} \frac{x^2}{n(n+x)} \\ & \leq \frac{x^2}{1(1+x)}+\sum_{n=2}^{\infty} \frac{x^2}{n(n+x)} \\ & \leq \frac{x^2}{1+x}+x^2 \sum_{n=2}^{\infty} \frac{x^2}{n(n+x)} \\ & \leq \frac{x^2}{1+x}+x^2 \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} \\ & \leq \frac{x^2}{1+x}+x^2\left(\frac{\pi^2}{6}=\varsigma(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}\right) \end{aligned}$$

Thus have we demonstrated the convergence of $\varphi(x)=\sum_{n=1}^{\infty} a_n(x)=\sum_{n=1}^{\infty}\left\{\frac{x}{n}-\ln \left(1+\frac{x}{n}\right)\right\}$.

As a special case:^[20]

$$\begin{aligned} \lim _{n \rightarrow \infty} \sum_{k=1}^n a_k(1) & =\lim _{n \rightarrow \infty} \sum_{k=1}^n\left\{\frac{1}{k}-\ln \left(1+\frac{1}{k}\right)\right\} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n\left\{\frac{1}{k}-\ln \left(\frac{k+1}{k}\right)\right\} \\ & =\lim _{n \rightarrow \infty}\left\{\sum_{k=1}^n \frac{1}{k}-\ln (n+1)\right\} \\ & =\lim _{n \rightarrow \infty}\left\{\sum_{k=1}^n \frac{1}{k}-\ln (n)\right\}=\gamma \approx 0.577 \end{aligned}$$

Returning to the limit in the log-gamma formula and replacing ’ $\alpha$ ’ with ’ $x$ ', we have:

$$\begin{aligned} & \lim _{n \rightarrow \infty}\left\{x \ln n-\sum_{k=1}^n \ln \left(1+\frac{x}{k}\right)\right\} \\ & \quad=\lim _{n \rightarrow \infty}\left\{x \ln n+\sum_{k=1}^n\left[-\frac{x}{k}+\frac{x}{k}-\ln \left(1+\frac{x}{k}\right)\right]\right\} \\ & \quad=\lim _{n \rightarrow \infty}\left\{x \ln n-\sum_{k=1}^n \frac{x}{k}\right\}+\sum_{k=1}^n\left\{\frac{x}{k}-\ln \left(1+\frac{x}{k}\right)\right\} \\ & \quad=-x \lim _{n \rightarrow \infty}\left\{\sum_{k=1}^n \frac{1}{k}-\ln (n)\right\}+\sum_{k=1}^n\left\{\frac{x}{k}-\ln \left(1+\frac{x}{k}\right)\right\} \\ & \quad=-\gamma x+\sum_{k=1}^n a_k(x) \end{aligned}$$

Thus, independently of the integral definition, we have proven for $\alpha>0$ the convergence of $\ln \Gamma(\alpha)=-\ln \alpha+\lim _{n \rightarrow \infty}\left\{\alpha \ln n-\sum_{k=1}^n \ln \left(1+\frac{\alpha}{k}\right)\right\}$.^[21]

This is but one example of a second-order convergence. To generalize, consider the convergence of $\varphi(x)=\lim _{n \rightarrow \infty} \sum_{k=1}^n f\left(\frac{x}{k}\right)$. As with the specific $f(x)=$ $x-\ln (1+x)$ above, the general function $f$ must be analytic over an open interval about zero, $a<0<b$. Likewise, its value and first derivative at zero must be analytic over an open interval about zero, $a<0<b$. Likewise, its value and first derivative at zero must be zero, i.e., $f(0)=f^{\prime}(0)=0$. According to the second-order mean value theorem, for $x \in(a, b)$, there exists a real $\xi$ between zero and $x$ such that:

$$f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(\xi)}{2} x^2=\frac{f^{\prime \prime}(\xi)}{2} x^2$$

Therefore, there exist $\xi_k$ between zero and $\frac{x}{k}$ such that $\varphi(x)=\lim _{n \rightarrow \infty} \sum_{k=1}^n f\left(\frac{x}{k}\right)=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{f^{\prime \prime}\left(\xi_k\right)}{2}\left(\frac{x}{k}\right)^2$.

Again, we appeal to absolute convergence. If $\lim _{n \rightarrow \infty} \sum_{k=1}^n\left|\frac{f^{\prime \prime}\left(\xi_k\right)}{2}\left(\frac{x}{k}\right)^2\right|=\frac{x^2}{2} \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{\left|f^{\prime \prime}\left(\xi_k\right)\right|}{k^2}$ converges, then $\varphi(x)$ converges. But every $\xi_k$ belongs to the closed interval $\Xi$ between 0 and $x$. And since $f$ is analytic over $(a, b) \supset \Xi, f^{\prime \prime}$ is continuous over $\Xi$. Since a continuous function over a closed interval is bounded, $0 \leq\left|f^{\prime \prime}\left(\xi_k\right)\right| \leq M$ for some real $M$. So, at length:

$$\begin{aligned} 0 \leq \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{\left|f^{\prime \prime}\left(\xi_k\right)\right|}{k^2} \leq \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{M}{k^2} & =M \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{k^2} \\ & =M \zeta(2)=M \frac{\pi^2}{6} \end{aligned}$$

Consequently, $\lim _{n \rightarrow \infty} \sum_{k=1}^n f\left(\frac{x}{k}\right)$ converges to a real number for every function $f$ that is analytic over some domain about zero and for which $f(0)=f^{\prime}(0)=0$. The convergence must be at least of the second order; a first-order convergence would involve $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{M}{k}$ and the divergence of the harmonic series.

The same holds true for analytic functions over complex domains. Another potentially important secondorder convergence is $\varphi(z)=\lim _{n \rightarrow \infty} \sum_{k=1}^n\left\{e^{\frac{z}{k}}-\left(1+\frac{z}{k}\right)\right\}$.