I want to prove that, for any $n \in \mathbb {Z}^+$, $$\ln(n + 1) \leq 1 + \frac 1 2 + … + \frac 1 n \leq 1 + \ln n\ ,$$ using the fact that, for any $i \in \mathbb {Z}^+$, $$\frac 1 {i + 1} \leq \ln(i + 1) – \ln i \leq \frac 1 i\ .$$
Some background
I am taking an introductory module to calculus in my university and this is actually one of the questions of my homework for this week. There are two parts to the question and this is the second part, which I have been stuck at for hours. The first part required me to prove that $$\frac 1 {i + 1} \leq \ln(i + 1) – \ln i \leq \frac 1 i\ ,$$ which I have finally been able to do. However, I am not getting any inspiration as to how I can use this to prove that $$\ln(n + 1) \leq 1 + \frac 1 2 + … + \frac 1 n \leq 1 + \ln n\ .$$
I did a little research online about the middle portion of the inequality and found that it is a special series called the harmonic series (no series are taught in my module). The Wikipedia page on the harmonic series also showed how to prove the lower and upper bounds that I am looking for, but they are far beyond my current ability.
Any intuitions/suggestions on how to carry on this problem will be greatly appreciated 🙂
Best Answer
For the second part, you in fact have two inequalities: \begin{align} \forall i \geqslant 1,~ \dfrac{1}{i+1} &\leqslant \ln(i+1) - \ln i &\text{and}~ \forall i \geqslant 1,~ \ln(i+1) - \ln i &\leqslant \dfrac{1}{i} \end{align} Sum the first for $i \in \{1,\ldots,n-1\}$, you find $\frac{1}{2} + \ldots + \frac{1}{n} \leqslant \ln n$. Add $1$ to both sides and you have \begin{align} \sum_{i=1}^n \frac{1}{n} \leqslant \ln n +1 \end{align}
Sum the second inequality for $i \in \{1,\ldots,n\}$ and you have \begin{align} \ln( n+1 ) \leqslant \sum_{i=1}^n \frac{1}{n} \end{align}