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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!

a9302c
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    There are different, but equivalent definitions of "logarithmic function". What is your definition of $\log$ in class? – Bumblebee Oct 26 '20 at 04:01
  • @Bumblebee The definition of "natural logarithmic function" (ln) given in class was ln(x) was equivalent to the definite integral of (1/t)dt from 1 to x. I was also introduced to the derivatives and integrals under the same section. – a9302c Oct 26 '20 at 04:24
  • Use the definition $\ln x =\int_1^x \frac 1t dt$. The result falls beautifully out. – fleablood Oct 26 '20 at 05:01
  • .... because for all $1 \le m\le t \le n$ then $\frac 1n\le \frac 1t \le \frac 1m$ so $\int_m^n \frac 1ndt \le \int_m^n \frac 1t dt \le \int_m^n\frac 1mdt$ – fleablood Oct 26 '20 at 05:13

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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.

Bumblebee
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  • Thank you for your answer. I am a little lost when you said you're integrating over the inequality to get the third equation. Can I clarify that you are letting $x = \frac{i+1}{i}$ and substituting the equation of the definition? – a9302c Oct 26 '20 at 07:00
  • @a9302c: I meant $\displaystyle\int_i^{i+1} \dfrac{1}{i+1}dt \le \displaystyle\int_i^{i+1} \dfrac{1}{t} dt \le \displaystyle\int_i^{i+1}\dfrac{1}{i}dt,$ and note that both left and right integrands are constant functions. – Bumblebee Oct 26 '20 at 17:04