-2

Please help me to prove this inequality

$$\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} < \ln n < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}$$

Dedhert . Jr
  • 17
  • 4
    You seem to have the inequality reversed. How can the leftmost expression be less than the rightmost? – saulspatz Oct 29 '20 at 14:29
  • 1
    @D Please note that askers are expected to provide context for their questions, as is explained here. For example, where did you find this question? What are your thoughts on the question? What have you tried so far? – Ben Grossmann Oct 29 '20 at 14:31
  • @Dedhert.Jr The first comment says that where you have written $a<b<c$, it should instead be $c < b< a$. – Ben Grossmann Oct 29 '20 at 14:33
  • In any case, just compare $\int_{k-1}^k\frac {dx} x$ to either $\frac 1{k - 1}$ or $\frac 1 k$. – WhatsUp Oct 29 '20 at 14:33
  • @RRL I disagree with the duplicate marking since the linked question does not directly ask about the same bounds. This post would be more apt – Ben Grossmann Oct 29 '20 at 14:35
  • 1
    @D I suspect, however, that you simply forgot the $1$ in front of the sum $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-1}$, which is a different issue – Ben Grossmann Oct 29 '20 at 14:36
  • @Ben Grosmann... So, was the question wrong? – Dedhert . Jr Oct 29 '20 at 14:37
  • 1
    @Dedhert If you have copied the question without any mistakes, then yes: the question must be wrong. We would have a better idea of what's going on, however, if you could provide context as I suggest in my first comment. – Ben Grossmann Oct 29 '20 at 14:39
  • Oh, thanks @Ben Grossmann. So, I actually found this question in one of the books, the name of books is Calculus, made by Varberg, Purcell, and Rigdon, Volume 1, Ninth edition. But it translated in Indonesian language. – Dedhert . Jr Oct 29 '20 at 14:42
  • @Dedhert.Jr Ok, thanks for explaining. Yes, it seems that there's a typo in the book then. I think that this is the question that the author meant to ask. – Ben Grossmann Oct 29 '20 at 15:04
  • Do you have a few questions that it similar to this question above? – Dedhert . Jr Oct 29 '20 at 15:40

2 Answers2

1

there is a simple picture for this,

if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} \; g(x) \; dx \; < \; \sum_{j=a}^b \; g(j) \; < \; \int_{a-1}^b \; g(x) \; dx $$

You need two versions, with $a=2,b=n$
$$ \sum_{j=2}^n \; \frac{1}{j} \; < \; \int_{1}^{n} \; \frac{1}{x} \; dx = \log n $$

$a=1 , b= n-1$ $$ \log n = \int_{1}^{n} \; \frac{1}{x} \; dx < \sum_{j=1}^{n-1} \; \frac{1}{j}$$

Will Jagy
  • 139,541
0

Hint:

$$\lfloor x\rfloor\le x\le\lceil x\rceil\implies \frac1{\lceil x\rceil}\le \frac1x\le \frac1{\lfloor x\rfloor}.$$

Integrate in unit intervals and sum.

$$\frac1n\le\log(n)-\log(n-1)\le\frac1{n-1}$$ which telescopes to $$H_{n}-1\le\log(n)-\log(1)\le H_{n-1}.$$