Is it always true that $\sum\limits_{j=1}^n\frac{1}{j} > \int_1^{n+1} \frac{dt}t$?

Question

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.

Answer 1

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.

Answer 2

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 $H_n>\log(n+1)$.