Does $c_{\ln(x)}=\lim_{n\to\infty}\left(\sum_{k=1}^n \ln(k)-\int_1^n \ln(x)\ dx\right)$ converge?

Question

Does $c_{\ln(x)}=\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n \ln(k)-\int_1^n \ln(x)\ dx\right)$ converge?

First of all, I always thought Mascheroni was spelled (and pronounced) Masechroni, but apparently this is wrong.

Anyway, the famous Euler-Mascheroni constant is defined as

$$\gamma = \displaystyle\lim_{n\to\infty}\left(-\ln n + \displaystyle\sum_{k=1}^n \frac1{k}\right).$$

Wikipedia states that many other constants arising from the difference of the sum and integral of a given function are known. But all the functions $f$ in this section are ones that converge as $\ x\to \infty.\ $ For example $f(x) = \frac{1}{x} \to 0\ $ as $x\to\infty.\ $ Continuing with the notation in that section, I am interested in:

$$\ c_{\ln(x)} = \displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n \ln(k) - \int_1^n \ln(x)\ dx\right) $$

Does the limit exist, and is the number known?

Or, for a function to give rise to such a constant, is it required that the function converges as $x\to\infty$?

The section on wikipedia does say, "for some arbitrary decreasing function $f$", which suggests, all one requires is that the function is decreasing: so in my question about $f(x) = \ln(x),\ $ we just look at $g(x) = -\ln(x)\ $ as this is decreasing, and the answer is "yes". Is this interpretation of what it says by "for some arbitrary decreasing function $f$" on wikipedia correct?

Edit: this cannot be what wikipedia means: for example, consider $f(x) = -x$.

So what does wikipedia mean here?

I guess a broader question would be: What conditions guarantee that a monotone function $f(x)$ have the property that $$\ c_{f} = \displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n f(k) - \int_1^n f(x)\ dx\right) $$

converges?

I know this is a lot of questions and it's meant to be limited to one question per question, but it's all closely related, so hopefully it's okay.

Answer 1

By Stirling's formula $$ \sum\limits_{k = 1}^n {\log k} = \log n! = \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \mathcal{O}\!\left( {\frac{1}{n}} \right). $$ Also $$ \int_1^n {\log x \,dx} = n\log n - n + 1. $$ Thus $$ \sum\limits_{k = 1}^n {\log k} - \int_1^n {\log x \, dx} = \frac{1}{2}\log n - 1 + \frac{1}{2}\log (2\pi ) + \mathcal{O}\!\left( {\frac{1}{n}} \right), $$ which tends to $+\infty$ as $n\to +\infty$. I recommend you to look up the Euler–Maclaurin formula and/or the Abel–Plana formula for the general case.

Answer 2

Very little is needed.

I got $$ \log k - \int_{k-1}^k \log x dx \approx \frac{1}{2k}. $$ Your expression is the sum of these pieces from $1$ to $n,$ and thereby diverges.

Answer 3

The question asks to find the area of the turquoise region.

Each turquoise shape is approximately a triangle, due to $\ln(x)$ being a "nice" (one with no crazy oscillations) function. More on this later.

Therefore,

$$\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n \ln(k)-\int_1^n \ln(x)\ dx\right) = \text{Area of turquoise region } $$

$$\approx \frac{1}{2} \left(\text{Area of upper rectangles (purple top) - area of lower rectangles (green top)}\right) $$

$$=\frac{1}{2}\displaystyle\lim_{n\to\infty}\left( \displaystyle\sum_{k=1}^n \ln(k+1)-\ln(k) \right) $$

$$ = \frac{1}{2}\displaystyle\lim_{n\to\infty}\left[\ (\color{red}{\ln(2)}-\ln(1))\ +\ (\ln(3)-\color{red}{\ln(2)})\ +\ \ldots\ +\ (\color{blue}{\ln(n)}-\ln(n-1))\ +\ (\ln(n+1)-\color{blue}{\ln(n)})\ \right] $$

$$ = \frac{1}{2}\displaystyle\lim_{n\to\infty}(\ \ln(n+1)-\ln(1)\ ) = \frac{1}{2}\times\infty = \infty. $$

$$$$

In fact, you can use the same idea to find an approximation of the Euler-Mascheroni constant itself, by applying the same method for $f(x)=\frac{1}{x}.$

$$\gamma = \displaystyle\lim_{n\to\infty}\left(-\ln n + \displaystyle\sum_{k=1}^n \frac1{k}\right)$$

$$ \approx\frac{1}{2}\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n \frac{1}{k}-\frac{1}{k+1}\right) $$

$$ =\frac{1}{2}\displaystyle\lim_{n\to\infty}\left(\ \left(1-\color{red}{\frac{1}{2}}\right)\ + \left(\color{red}{\frac{1}{2}}-\frac{1}{3}\right)\ + \ldots + \left(\frac{1}{n-1}-\color{blue}{\frac{1}{n}}\right)\ + \left(\color{blue}{\frac{1}{n}}-\frac{1}{n+1}\right)\ \right)$$

$$ =\frac{1}{2}\displaystyle\lim_{n\to\infty}\left( 1- \frac{1}{n+1} \right)=\frac{1}{2},$$

corresponding to the fact that the Euler-Mascheroni constant, $\gamma = 0.577...,\ $ which is not far from $\frac{1}{2}.$

I think this also answers the question functions in general, by applying the same cancellation technique. Thus,

The limit

$$\ c_{f} = \displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n f(k) - \int_1^n f(x)\ dx\right) $$

converges if and only if $\ \displaystyle\lim_{x\to\infty}f(x)\ $ exists and is finite.