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This inequality was presented as obvious in this answer. I was interested in understanding the details for this comparison so I started reading up the Euler-MacLaurin formula but this actually confused me since computation of the remainder is very detailed (I have not spent enough time working with Bernoulli Numbers to have an intuition on their range of values).

I wasn't clear if the use in the answer above presupposed the conditions identified in the question or whether it was always true.

If it is always true, if someone can outline a simple argument that would be great. I have seen a visual argument but I am not clear how this translates to a proof for all integer ranges.

Kamal Saleh
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Larry Freeman
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  • It is true and it has been used many many times on this site. Just split the integral on the right into integrals from $j$ to $j+1, 1 \leq j \leq n$. – Kavi Rama Murthy Jan 05 '23 at 23:31
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    Crucially, $f(t) = \frac1t$ is positive and decreasing so the Riemann sum on $[1, n+1]$ with rectangles of width $1$, sampled on the left-hand side of each subinterval will always overestimate the integral. – Sammy Black Jan 05 '23 at 23:32
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    if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} ; g(x) ; dx ; < ; \sum_{j=a}^b ; g(j) ; < ; \int_{a-1}^b ; g(x) ; dx $$

    $$ \int_2^{n+1} ; \frac{1}{x} ; dx ; < ; \sum_{j=2}^n ; \frac{1}{j} ; < ; \int_{1}^n ; \frac{1}{x} ; dx $$ $$ 1 +\int_2^{n+1} ; \frac{1}{x} ; dx ; < ; 1 + \sum_{j=2}^n ; \frac{1}{j} ; < ; 1 + \int_{1}^n ; \frac{1}{x} ; dx $$

    $$ 1 +\int_2^{n+1} ; \frac{1}{x} ; dx ; < ; \sum_{j=1}^n ; \frac{1}{j} ; < ; 1 + \int_{1}^n ; \frac{1}{x} ; dx $$

     – Will Jagy Jan 05 '23 at 23:40
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    well, see earlier today https://math.stackexchange.com/questions/4612492/logn-frac1n-h-n-logn-1 – Will Jagy Jan 06 '23 at 01:29
  • Concerning the visual argument see for example here. – Raymond Manzoni Jan 13 '23 at 16:52

2 Answers2

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On $(k,k+1)$, $1/x<1/k$, so

$$\int_1^{n+1} \frac{1}{x} dx = \sum_{k=1}^n \int_k^{k+1} \frac{1}{x} dx < \sum_{k=1}^n \frac{1}{k} < \sum_{k=1}^{n+1} \frac{1}{k}.$$

The main reason we care about $n+1$ is not for this bound (which as I showed here does not need to involve a sum all the way to $n+1$). Rather, it is for the other direction, where you take right hand sums. For that, you say that on $(k,k+1),1/x>1/(k+1)$, so

$$\int_1^{n+1} \frac{1}{x} dx = \sum_{k=1}^n \int_k^{k+1} \frac{1}{x} dx > \sum_{k=1}^n \frac{1}{k+1} = \sum_{k=2}^{n+1} \frac{1}{k}.$$

Both of these are basically just the "visual argument" you mentioned written down in symbols.

Ian
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It is known that $$H_n=\sum_{k=1}^n\frac{1}{k}>\log n +\frac{1}{n}=\log(ne^{1/n})>\log(n+1)$$ from this question that was asked recently. It is also known that $$\lim_{n\rightarrow\infty}H_n-\log(n+a)=\gamma>0$$For any number $a$, where $\gamma$ is Euler's constant. This shows that it is not just $\log(n+1)$ that is less than $H_n$, but so as $\log(n+a)$ for all $a$ at some $n$. In this case, $H_n=\log(n+1)$ when $n=0$ and so $H_n>\log(n+1)$.

Kamal Saleh
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