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I have the following limit

$$ \lim_{x \to \infty} \frac{{\rm d}}{{\rm d}x} \left( \sum_{k=0}^x \frac{1}{k} \right) $$

Since I am taking the limit only after I take the derivative of this series, I get

$$ \lim_{x \to \infty} \frac{{\rm d}}{{\rm d}x} \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{x} \right) = \lim_{x \to \infty} \left(- \frac{1}{x^2} \right) = 0^- $$

Even though we are taking the limit to infinity after the derivative, if I were to put it before, the series would diverge and then it would be completely different. Is anything wrong with what I did?

I ask this because I'm not entirely sure if I can interchange the limit and the derivative at will, but from this example I clearly cannot.

EDIT:

$$ \lim_{x \to \infty } \left( \frac{\ln(x)}{\Gamma(x+1)} \right) = ? $$

Rodrigo de Azevedo
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MicrosoftBruh
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    what does $\sum_{k=0}^{7/2}\frac{1}{k}$ mean? – peek-a-boo Apr 06 '22 at 01:34
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    Forget about limits and sums, and exchanging derivatives with limits and all these analysis pathologies. The very first step is you need to have a well-defined function on an interval $(0,\infty)$ in order to even talk about the operations you're trying to do. That's the issue here. – peek-a-boo Apr 06 '22 at 01:35
  • So, limits to infinity involving factorials for instance would have to be dealt with using the gamma function, correct? But if I use L'Hopital and stumble upon a derivative of said factorial I would get the series in question (apart from the Euler-Mascheroni constant) – MicrosoftBruh Apr 06 '22 at 01:48
  • I can edit the question and put the limit I'm talking about. – MicrosoftBruh Apr 06 '22 at 01:48
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    We can certainly talk about limits of factorials without the Gamma function; though this has to be treated as a limit of a sequence of numbers, not a function defined on $\Bbb{R}$. We can't however talk about derivatives of factorials. THe factorial $!$ is only defined for non-negative integers, by recursion $0!:=1$, and for $n\geq 0$, we define $(n+1)!:=(n+1)\cdot n!$. – peek-a-boo Apr 06 '22 at 01:49
  • I saw the derivative of the factorial here: https://math.stackexchange.com/questions/300526/derivative-of-a-factorial – MicrosoftBruh Apr 06 '22 at 01:51
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    No, what you see in the top answer there is not the derivative of the factorial. That makes no sense (derivatives are defined for functions defined on open intervals, they make no sense for functions defined on the natural numbers). WHat you see in that link is the derivative of the Gamma function. These are separate things. It is only by abuse of language that people associate factorials with the Gamma function. – peek-a-boo Apr 06 '22 at 01:53
  • Okay, thank you. – MicrosoftBruh Apr 06 '22 at 01:57
  • @peek-a-boo How about now? That series would appear if I used L'Hopital on that limit, how would I deal with that? – MicrosoftBruh Apr 06 '22 at 02:11
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    I can tell you $\lim\limits_{x\to \infty}\frac{\ln(x)}{\Gamma(x+1)}=0$, but still it has no relation to your original (undefined) sum. L'Hopital's rule is a theorem about differentiable functions and you still haven't provided a definition for what you're talking about in the first line. – peek-a-boo Apr 06 '22 at 02:20
  • @peek-a-boo Can you show me what you did to evaluate that limit? – MicrosoftBruh Apr 06 '22 at 11:48
  • I don't understand that which comes after EDIT. – Rodrigo de Azevedo Apr 10 '22 at 11:46
  • @RodrigodeAzevedo I ultimately wanted to solve the limit that is written on the EDIT part, but what I didn't understand was that particular step of the derivative of the series, since it appeared in my attempt. So I just asked the series part because I think I can handle the rest. – MicrosoftBruh Apr 10 '22 at 20:42
  • @MicrosoftBruh Then please include such information in the question itself. – Rodrigo de Azevedo Apr 11 '22 at 04:39

2 Answers2

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$$\lim_{x \to \infty} \frac{{\rm d}}{{\rm d}x} \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{x} \right) = \lim_{x \to \infty} \left(- \frac{1}{x^2} \right) $$

This arguement is false because: \begin{eqnarray} \require{cancel} \dfrac{{\rm d}}{{\rm d}x}\left(\sum_{k=1}^{x}{\dfrac{1}{k}}\right)&=&\lim_{h\to 0}{\dfrac{\sum_{k=1}^{x+\color{red}{h}}{\dfrac{1}{k}}-\sum_{k=1}^{x}{\dfrac{1}{k}}}{h}}\\&=&\lim_{h\to 0}{\dfrac{\sum_{k=x+1}^{x+h}{\dfrac{1}{k}}}{h}}\\ &=&\lim_{h\to 0}{\dfrac{\sum_{k=x+1}^{x+1+h}{\dfrac{1}{k}}-\dfrac{1}{x+1+h}}{h}}\\ &=&\cancelto{\dfrac{-1}{0}}{\lim_{h\to 0}{\dfrac{(x+1+h)\sum_{k=x+1}^{x+1+h}{\dfrac{1}{k}}-1}{h(x+1+h)}}}\\ &\neq&\dfrac{-1}{x^2}\\ \end{eqnarray} The "correct" limit does not exist because this isn't it since we assumed $h$ is an integer when it's not.

Thus, we can't take the derivative via its definition.

However, there is another way around; the digamma function allows us to generate harmonic numbers and solve for the limit: \begin{eqnarray} \sum_{k=1}^{x}{\dfrac{1}{k}}&=&H_x\\ \psi(x+1)&=&\dfrac{\Gamma'(x+1)}{\Gamma(x+1)}\\ \Gamma(x+1)&=&x \Gamma(x)\\ \Gamma'(x+1)&=&\Gamma(x)+x\Gamma'(x)\\ \psi(x+1)&=&\dfrac{\Gamma(x)+x \Gamma'(x)}{x\Gamma(x)}\\ &=&\dfrac{1}{x}+\dfrac{\Gamma'(x)}{\Gamma(x)}\\&=&\dfrac{1}{x}+\psi(x)\\ &=&\psi(2-1)+\sum_{k=1}^{x}{\dfrac{1}{k}}\\ &=&\psi(1)+H_x\\ \Gamma(x+1)&=&\int_{0}^{\infty}{t^xe^{-t}dt}\\ \Gamma'(x+1)&=&\dfrac{{\rm d}}{{\rm d}x}\int_{0}^{\infty}{t^xe^{-t}dt}\\ &=&\int_{0}^{\infty}{\dfrac{ \partial}{\partial x}\left(t^xe^{-t}\right)dt}\tag{*}\\ &=&\int_{0}^{\infty}{t^xe^{-t}\ln(t)dt}\\ \lim_{x\to \infty}{\ln\left(1-\dfrac{t}{x}\right)^x}&=&\lim_{x\to \infty}{x\ln\left(1-\dfrac{t}{x}\right)}\\ u&=&\dfrac{1}{x}\\ x\to \infty&\Leftrightarrow&u\to0^+\\ \lim_{x\to \infty}{x\ln\left(1-\dfrac{t}{x}\right)}&=&\lim_{u\to 0^+}{\dfrac{\ln\left(1-ut\right)}{u}}\\ &\overset{\tiny\dfrac{0}{0}}{=}&\lim_{u\to 0^+}{\dfrac{-t}{1-ut}}\\ &=&-t\\ \lim_{x\to \infty}{\left(1-\dfrac{t}{x}\right)^{x-1}}&=&\lim_{x\to \infty}{\exp\left(\ln\left(1-\dfrac{t}{x}\right)^x\right)\left(1-\dfrac{t}{x}\right)^{-1}}\\ &=&e^{-t}\\ \int_{0}^{\infty}{e^{-x}\ln(t)dt}&=&\int_{0}^{\infty}{\lim_{x\to \infty}{\left(1-\dfrac{t}{x}\right)^{x-1}}\ln(t)dt}\\ &=&\lim_{x\to \infty}{\int_{0}^{x}{{\left(1-\dfrac{t}{x}\right)^{x-1}}\ln(t)dt}}\tag{**}\\ u&=&1-\dfrac{t}{x}\\ t&=&x(1-u)\\ dt&=&-xdu\\ t\to 0&\Leftrightarrow&\,u\to 1\\ t\to x&\Leftrightarrow&\,u\to 0\\ \lim_{x\to \infty}{\int_{0}^{x}{{\left(1-\dfrac{t}{x}\right)^{x-1}}\ln(t)dt}}&=&\lim_{x\to \infty}{-x\int_{1}^{0}{{u^{x-1}}\ln(x(1-u))du}}\\ &=&\lim_{x\to \infty}{x\int_{0}^{1}{{u^{x-1}}\ln(x(1-u))du}}\\ &=&\lim_{x\to\infty}{\left(x\int_{0}^{1}\ln(x)u^{x-1}du+x\int_{0}^{1}u^{x-1}\ln(1-u)du\right)}\\ &=&\lim_{x\to\infty}{\left(\ln(x)+x\int_{0}^{1}u^{x-1}\ln(1-u)du\right)}\\ \operatorname{G}_u&=&1+u+u^2+u^3+\cdots\\ &=&1+u(1+u^2+u^3+\cdots)\\ &=&1+u\operatorname{G}_u\\ &=&\dfrac{1}{1-u} \Leftrightarrow|u|< 1\\ \dfrac{{\rm d}(\ln(1-u))}{{\rm d}u}&=&\dfrac{-1}{1-u}\\ &=&-(1+u+u^2+\cdots)\Leftrightarrow{|u|<1\Leftarrow u\in (0,1)}\\ \ln(1-u)&=&-\int(1+u+u^2+\cdots)du\\ &=&-\left(\dfrac{u}{1}+\dfrac{u^2}{2}+\dfrac{u^3}{3}+\cdots\right)\\ \lim_{x\to\infty}{\left(\ln(x)+\int_{0}^{1}u^{x-1}\ln(1-u)du\right)}&=&\lim_{x\to\infty}{\left(\ln(x)-x\int_{0}^{1}u^{x-1}\left(\dfrac{u}{1}+\dfrac{u^2}{2}+\dfrac{u^3}{3}+\cdots\right)du\right)}\\ &=&\lim_{x\to\infty}{\left(\ln(x)-x\left[\dfrac{u^{x+1}}{1(x+1)}+\dfrac{u^{x+2}}{2(x+2)}+\cdots\right]_{0}^{1}\right)}\\ &=&\lim_{x\to\infty}{\left(\ln(x)-\sum_{k=1}^{\infty}{\dfrac{x}{k(x+k)}}\right)}\\ &=&\lim_{x\to\infty}{\left(\ln(x)+\sum_{k=1}^{\infty}{\dfrac{1}{k}}-\sum_{k=1}^{\infty}{\dfrac{1}{x+k}}\right)}\\ &=&\lim_{x\to\infty}{\left(\ln(x)-\sum_{k=1}^{x}{\dfrac{1}{k}}\right)}\\ \int_{0}^{\infty}{e^{-t}\ln(t)dt}&=&-\lim_{x\to\infty}{\left(-\ln(x)+\sum_{k=1}^{x}{\dfrac{1}{k}}\right)}\\ \Gamma'(1)&=&-\gamma\\ \psi(1)&=&\dfrac{\Gamma'(1)}{\Gamma(1)}\\ &=&\Gamma'(1)\\ &=&-\gamma\\ \psi(x+1)&=&\psi(1)+H_x\\ H_x&=&\gamma+\dfrac{\int_{0}^{\infty}{t^xe^{-t}\ln(t)dt}}{\int_{0}^{\infty}t^xe^{-t}dt}\\ H^{'}_x&=&\dfrac{1}{\left(\int_{0}^{\infty}{t^xe^{-t}dt}\right)^2}\left(\int_{0}^{\infty}{t^xe^{-t}\ln^2(t)dt} \int_{0}^{\infty}{t^xe^{-t}dt}-\left(\int_{0}^{\infty}{t^xe^{-t}\ln(t)dt}\right)^2\right)\\ &=&\dfrac{\int_{0}^{\infty}t^xe^{-t}\ln^2(t)dt}{\int_{0}^{\infty}t^xe^{-t}dt}-\left(\dfrac{\int_{0}^{\infty}t^xe^{-t}\ln(t)dt}{\int_{0}^{\infty}t^xe^{-t}dt}\right)^2\\ \lim_{x\to\infty}{H^{'}_{x}}&=&\lim_{x\to\infty}{\dfrac{\int_{0}^{\infty}t^xe^{-t}\ln^2(t)dt}{\int_{0}^{\infty}t^xe^{-t}dt}}-\left(\lim_{x\to\infty}{\dfrac{\int_{0}^{\infty}t^xe^{-t}\ln(t)dt}{\int_{0}^{\infty}{t^xe^{-t}dt}}}\right)^2\\ &\overset{\tiny\dfrac{\infty}{\infty}}{=}&\lim_{x\to\infty}{\dfrac{\int_{0}^{\infty}{t^xe^{-t}\ln^3(t)dt}}{\int_{0}^{\infty}{t^xe^{-t}\ln(t)dt}}}-\left(\lim_{x\to\infty}{\dfrac{\int_{0}^{\infty}{t^xe^{-t}\ln^2(t)dt}}{\int_{0}^{\infty}{t^xe^{-t}\ln(t)dt}}}\right)^2\\ \\ &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\cdot\\ &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\cdot\\ &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\cdot\\ \\ &\overset{\tiny\dfrac{\infty}{\infty}}{=}&\lim_{x\to\infty}{\lim_{n\to\infty}{\dfrac{\int_{0}^{\infty}t^xe^{-t}\ln^{n+2}(t)dt}{\int_{0}^{\infty}{t^xe^{-t}\ln^{n}(t)dt}}}}-\left(\lim_{x\to\infty}{\lim_{n\to\infty}{\dfrac{\int_{0}^{\infty}{t^xe^{-t}\ln^{n+1}(t)dt}}{\int_{0}^{\infty}{t^xe^{-t}\ln^{n}(t)dt}}}}\right)^2\\ &=&1-1^2\\ \lim_{x\to\infty}{\dfrac{{\rm d}}{{\rm d}x}\left(\sum_{k=1}^{x}{\dfrac{1}{k}}\right)}&=&0 \end{eqnarray} Where $\Gamma(x)$ is the gamma function, $\psi(x)$ is the digamma function and $\gamma$ is the Euler-mascheroni constant. (*) uses Leibnitz' rule of integration. (**) uses the dominated convergence theorem.

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If you just want to find: $$\lim_{x\to \infty}{\dfrac{\ln x}{\Gamma(x+1)}} $$ then it's pretty easy.

\begin{eqnarray} \lim_{x\to \infty}{\dfrac{\ln x}{\Gamma(x+1)}}&=&\lim_{x\to\infty}{\dfrac{\ln x}{x\Gamma(x)}}\\ &\overset{\tiny\dfrac{\infty}{\infty}}{=}&\lim_{x\to \infty}{\dfrac{\dfrac{1}{x}}{\Gamma(x)+x\Gamma'(x)}}\\ &=&\lim_{x\to \infty}{\dfrac{1}{x\Gamma(x)+x^2\Gamma'(x)}}\\ &=&0\\ \end{eqnarray} By: \begin{eqnarray} \lim_{x\to\infty}{\dfrac{\Gamma(x)}{x-1}}&=&\lim_{x\to \infty}{\dfrac{(x-1)\Gamma(x-1)}{x-1}}\\ &=&\lim_{x\to \infty}{\Gamma(x-1)}\\ &=&\infty\\ \lim_{x\to\infty}{\dfrac{\Gamma(x)}{x-1}}&\overset{\tiny\dfrac{\infty}{\infty}}{=}&\lim_{x\to \infty}{\Gamma'(x)}\\ \lim_{x\to\infty}{\Gamma'(x)}&=&\infty \end{eqnarray}