$\sum_{x=a}^{b-1}\frac{1}{x}$ and $\sum_{x=a+1}^b\frac{1}{x}$

Question

I have to prove the following relations:

• $\sum_{x=a}^{b-1}\frac{1}{x}\geq\log b - \log a $

• $\sum_{x=a+1}^{b}\frac{1}{x}\leq\log b - \log a $

I tried to use the relation that $\int_a^b \frac{1}{x} dx = \log b - \log a$ and for the second relation I obtained that $\int_{a+1}^b \frac{1}{x} dx = \log b - \log (a+1) < \log b - \log a$ and I think that could be sufficient to prove the relation. Instead, for the first relation i have some problems, because using the same method I don't obtain a good result, indeed: $ \int_{a}^{b-1} \frac{1}{x} dx = \log (b-1) - \log a$ that is not greater than $\log b - \log a$. What do you think? How can I resolve it?

Answer 1

First, welcome to MSE, and great opening post!

Here's my hints on how to analyze this problem:

Consider left hand and right hand sums of $\frac{1}{x}$. For our first inequality we are using lefthand sums with width of $1$, so then, by the shape of the graph $\frac{1}{x}$ we get an overestimation of the integral so that $$ \sum_{x=a}^{b-1} \frac{1}{x} \ge \int_a^b \frac{1}{x} \, \mathrm{d} x $$

Do you see what to do for the second example?

As for your approach to the problem, you were close, realizing that $\int_{a+1}^b \frac{1}{x} \, \mathrm{d} x \le \int_a^b \frac{1}{x} \, \mathrm{d} x$ is similar to what I'm doing - it just seems like you were one step away!