Prove $\int_{1}^{x} \frac{dt}{t} \leq \sum_{n \leq x} \frac{1}{n} \leq 1 + \int_{1}^{x} \frac{dt}{t}$

Question

Let $x \geq 1$. I wish to show, by interpreting the sum as a Riemann sum, that $$\log(x) = \int_{1}^{x} \frac{dt}{t} \leq \sum_{n \leq x} \frac{1}{n} \leq 1 + \int_{1}^{x} \frac{dt}{t}.$$ Certainly, $f(t) = 1/t$ is continuous on $[1, x]$, so is integrable on $[1, x]$ and may be computed as the limit of a sequence of Riemann sums. I have only the idea of interpreting $\sum_{n \leq x} \frac{1}{n}$ as an "upper sum" of $f$ with respect to a partition $\mathcal{P} = \{1, 2, ..., \lfloor x \rfloor, x\}$, which would necessarily be no less than $\int_{1}^{x} \frac{dt}{t}$. However, I am not certain how to formalise this idea, because, in using $\mathcal{P}$, I should be left with an "excess" area from $\lfloor x \rfloor$ to $x$?

Answer 1

It would be convenient, since the sum is divided up at the integers, to interpret your Riemann sum as having equidistant points $1$ unit of distance apart each. So:

$$ \int_1^x \frac{dt}{t} = \sum_{n=1}^{\lfloor x \rfloor -1} \int_n^{n+1} \frac{dt}{t} + \int_{\lfloor x \rfloor}^x \frac{dt}{t} $$

Since $\frac 1 t$ is decreasing, then $$ \frac{1}{n+1} \le \frac 1 t \le \frac{1}{n} \text{ on } [n,n+1] $$

Then $$ \frac{1}{n+1} \le \int_n^{n+1} \frac{dt}{t} \le \frac 1 n $$

These are best easily seen as in, say, this Desmos demo. (Modified from one I use to show students about the left-endpoint rule.)

From here, the remainder is ultimately just careful manipulations & algebra.