TITLE: Solving a question with the definition of logarithmic function
QUESTION [1 upvotes]: For my calculus class, I have been given a question to solve using the "definition of the logarithmic function", to prove that for any $i\in\mathbb{Z}^+$,
$\frac{1}{i+1} \leq ln(i+1) - ln(i) \leq \frac{1}{i}$
then go on to prove using the previous result that
$ln(n+1) \leq 1 + \frac{1}{2} + ...+ \frac{1}{n} \leq 1 + ln(n)$
I understand decently the properties of the logarithmic function, its derivatives and integrals (slightly shaky on the last one), but I'm completely lost on what the question is asking about. I've tried to differentiate
$ln(i+1) - ln(i)$ to get $\frac{1}{i+1} - \frac{1}{i}$ which corresponds with the two fractions in the conditions but I'm stuck from here onwards.
Can someone offer me some direction/advice on how I can prove the question?
Thank you!

REPLY [1 votes]: Hint: Let me assume your definition of natural logarithm is $$\ln x=\displaystyle\int_1^x\dfrac{1}{t}dt.$$ Lets consider the interval $[i, i+1]$ where $i\in\mathbb{Z}^+$ as given in your problem statement. Then note that $$\dfrac{1}{i+1}\le\dfrac{1}{t}\le\dfrac{1}{i}$$ for any $t$ in the interval. Then integration of this inequality over the interval to obtain the inequality $$\dfrac{1}{i+1} \leq \ln(i+1) - \ln(i) \leq \dfrac{1}{i}.$$ Take the summation of such inequalities from $i=1$ to $i=n-1$ (this is called a telescoping summation). It proves that $$\sum_{i=2}^n\dfrac{1}{i}\le \ln n\le \sum_{i=1}^{n-1}\dfrac{1}{i}$$ and hence your last required inequality.
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