1

Prove that

$$ \sum_{k = 1}^{n}\frac{1}{k} - \ln\left(n\right) \to \int_{0}^{1}\left[\frac{1}{1 - x} + \frac{1}{\ln\left(x\right)}\right]{\rm d}x \quad\mbox{as}\quad n\to\infty $$

I don't think this question is a duplicate of this post, because I'm looking for a more direct method. I also am not sure why the result from that post implies the result for this question. Here is the original source of this question.

I think it could be useful to substitute $x=e^{-t}$. Then $$ \int_{0}^{1}\left[\frac{1}{1 - x} + \frac{1}{\ln\left(x\right)}\right]{\rm d}x = \int_{\infty}^{0}\left({1 \over 1-e^{-t}} - {1 \over t}\right) \left(-{\rm e}^{-t}\right){\rm d}t = \int_{0}^{\infty}\left({{\rm e}^{-t} \over 1-{\rm e}^{-t}} + {{\rm e}^{-t} \over t}\right){\rm d}t $$ We also know that $$ \dfrac{e^{-t}}{t} = \sum_{n=1}^\infty\left[\,{\dfrac{e^{-nt}}{t} - \dfrac{e^{-(n+1)t}}{t}}\,\right]\quad\mbox{and}\quad \dfrac{e^{-t}}{1-e^{-t}} = \sum_{n=1}^\infty e^{-nt} $$ Thus, assuming we can interchange sums and integrals, we have $$ \int_{0}^{\infty}\left({{\rm e}^{-t} \over 1-e^{-t}} + {e^{-t} \over t}\right){\rm d}t = \sum_{n=1}^\infty \int_0^\infty \left[\,{\dfrac{e^{-nt}}{t} - \dfrac{e^{-(n+1)t}}{t}}\,\right]{\rm d}t + \dfrac{1}{n} $$ But I'm not sure how to evaluate $\int_{0}^{\infty}{\rm e}^{-nt}/t\,{\rm d}t $. Also, does the latter integral even exist $?$. I think I could use integration by parts or use the Taylor expansion of ${\rm e}$, but if I use the latter expansion, then I'm not sure how to justify interchanging integrals.

Edit: I'm looking for solutions that are elementary enough for me to understand. Using the DCT is fine since I've seen a proof of that and I've learned some basic measure theory. I don't currently fully understand any of the solutions in the proposed duplicate, so they're not elementary enough.

Edit #2: According to the solution below, it seems that $\int_{0}^{1}\dfrac{x^{n - 1} - 1}{\ln\left(x\right)}{\rm d}x = \ln\left(n\right)$. I was wondering how to prove this.

Felix Marin
  • 89,464
user33096
  • 2,021
  • Check this: https://math.stackexchange.com/a/980715/42969 – Martin R Dec 05 '22 at 13:14
  • It is also one of the identities here: https://en.wikipedia.org/wiki/Euler%27s_constant#Integrals – Martin R Dec 05 '22 at 13:17
  • You cannot compute $\int_0^\infty\mathrm e^{-nt}\frac{\mathrm dt}t$ and $\int_0^\infty\mathrm e^{-(n+1)t}\frac{\mathrm dt}t$ separately because the two integrals compensate (you have $\infty-\infty$). But one could first show that $\int_0^\infty\left(\mathrm e^{-t}-\mathrm e^{-at}\right)!\frac{\mathrm dt}t=\log a$ for any $a>0$, so that $$\int_0^\infty\left(\frac{\mathrm e^{-nt}}t-\frac{\mathrm e^{-(n+1)t}}t\right)\mathrm dt=\log(n+1)-\log n.$$ – nejimban Dec 05 '22 at 13:18
  • Use Feymans's trick to to show. – xpaul Dec 05 '22 at 20:28

1 Answers1

2

In fact \begin{eqnarray} &&\int_0^1 \bigg(\frac1{1-x} + \frac1{\ln x}\bigg) dx \\ &=&\int_0^1 \bigg(\frac{1-x^n}{1-x} -\frac{x^{n-1}-1}{\ln x}\bigg)dx+\int_0^1x^{n-1}\bigg(\frac{x}{1-x}+\frac1{\ln x}\bigg) dx \\ &=&\sum_{k=1}^n \frac1k - \ln n +\int_0^1x^{n-1}\bigg(\frac{x}{1-x}+\frac1{\ln x}\bigg) dx. \end{eqnarray} Noting that $$ \bigg|x^{n-1}\bigg(\frac{x}{1-x}+\frac1{\ln x}\bigg)\bigg|\le \bigg|\frac{x}{1-x}+\frac1{\ln x}\bigg|,\\ \frac{x}{1-x}+\frac1{\ln x}=\frac{x\ln x+1-x}{(1-x)\ln x}<0 \text{ for }x\in(0,1) $$ and $$ \int_0^1 \bigg(\frac1{1-x} + \frac1{\ln x}\bigg) dx $$ exists, by DCT, $$ \lim_{n\to\infty}\int_0^1x^{n-1}\bigg(\frac{x}{1-x}+\frac1{\ln x}\bigg) dx=0$$ which implies one has $$ \lim_{n\to\infty}\sum_{k=1}^n \frac1k - \ln n=\int_0^1 \bigg(\frac1{1-x} + \frac1{\ln x}\bigg) dx. $$

xpaul
  • 44,000
  • Thanks. I know about the DCT and I've seen a proof before. Just to justify some of your steps a bit further, we can alternatively argue that $\dfrac{x\ln x + 1-x}{(1-x)\ln x}$ is a continuous function of x for $x\in (0,1)$, and using L'hopital's rule, we see that its limit at $0$ from above is $0$ while its limit at 1 from below is $-1/2$. Hence for x sufficiently close to 1, it is close to $-1/2$ and it is sufficiently close to $0$ when x is sufficiently close to 0. The EVT can then be used to obtain the uniform bound. Could you see my edit in my question? – user33096 Dec 05 '22 at 19:12
  • It seems the proof of the latter equality can be shown using Leibniz's criterion. The derivative with respect to n of the LHS is just $\int_0^1 x^{n-1} dx = \dfrac{1}n,$ and integrating with respect to n and noticing that for $n=1,$ both functions equal zero, we see that the desired equality holds. – user33096 Dec 05 '22 at 23:02