Proving that sequence is decreasing

Question

I was looking at the first answer to this question Simple proof of showing the Harmonic number $H_n = \Theta (\log n)$, which claims that the following sequence is decreasing: $$ f(n) = H_n - \log(n) $$ where $\log$ is the natural log and $H_n = \sum_{i = 1}^n \frac{1}{i}$.

Here is my attempt to verify this:

Observe that $$ f(n + 1) - f(n) = H_{n + 1} - \log(n + 1) - (H_n - \log(n)) = \frac{1}{n + 1} + \log \left(\frac{n}{n + 1} \right). $$ But this last term is positive, so it appears that the sequence is increasing! Where is my mistake?

Welcome to MSE. Note that although $\frac{1}{n + 1}$ is positive, $\log \left(\frac{n}{n+1}\right)$ is negative as $\frac{n}{n+1} \lt 1$. — John Omielan, Jan 22 '19 at 01:16
no, $n/(n+1)$ is smaller than one, so its logarithm is negative. — Will Jagy, Jan 22 '19 at 01:17
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative? — Submarine, Jan 22 '19 at 01:18
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion. — John Omielan, Jan 22 '19 at 01:24
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative? — Submarine, Jan 22 '19 at 01:26

score 2 · Answer 1 · answered Jan 22 '19 at 01:29

2

Hint:

$$\frac{1}{n+1} < \int_n^{n+1} \frac{dt}{t} = \log (n+1) - \log n \\ \implies \frac{1}{n+1} -\log (n+1) < - \log n$$

answered Jan 22 '19 at 01:29

RRL

90,707

score 0 · Answer 2 · answered Jan 22 '19 at 02:00

As the other comments noted, although $\frac{n}{n+1}$ is positive, it is also less than 1 so its natural logarithm must be negative. Thus, the problem comes down to showing that $$\frac{1}{x} + ln\Big(\frac{x-1}{x}\Big)<0$$ That is equivalent to $$f(x)=\frac{1}{x} - ln\Big(\frac{x}{x-1}\Big) < 0.$$ It's easy to see that $$\lim_{x \to \infty} f(x) = 0$$ and also that $$f^\prime (x) = \frac{1}{x^2(x-1)},$$ which is positive for all $x<1$. Thus, $f$ is a strictly inscreasing function. As such, $max(f(x))=\lim_{x\to\infty}f(x)=0$.

Additionally, $f$ is not defined at $x=0$ so $f$ must be strictly negative.

Proving that sequence is decreasing

2 Answers2