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Let $N\geqslant1$ be an integer. Then which of the following statements are true ?

  1. $\displaystyle\sum\limits_{n=1}^N\dfrac1n\leqslant1+\log N$

  2. $\displaystyle\sum\limits_{n=1}^N\dfrac1n<1+\log N$

  3. $\displaystyle\sum\limits_{n=1}^N\dfrac1n\leqslant\log N$

  4. $\displaystyle\sum\limits_{n=1}^N\dfrac1n\geqslant\log N$

My Attempt : $$\underbrace{\frac11+\frac11+\frac11+\frac11+\frac11+\frac11+\ldots+\frac11}_{N\text{ times}}=N$$

$$\geqslant\sum_{n=1}^N\frac1n=\frac11+\frac12+\frac13+\frac14+\frac15+\frac16+\ldots+\frac1N$$

$$\geqslant\underbrace{\frac1N+\frac1N+\frac1N+\frac1N+\frac1N+\frac1N+\ldots+\frac1N}_{N\text{ times}}$$

$$=\frac NN=1$$

$$\implies1\leqslant\sum_{n=1}^N\frac1n\leqslant N$$

Also $\;\log N\leqslant N,\;\forall N\geqslant1\,.\;$ I'm unable to combine this information to answer the question. Please help me.

Angelo
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Mera bhai
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2 Answers2

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$$\log(1+x)<x\;,\;\;\text{ if }x>0$$ Let $\;x=\dfrac1n\;,\;$ then $$\frac1n>\log(1+n)-\log n$$ Next, by telescopic summing we get $$\sum_{n=1}^N\frac1n>\log(1+N)-\log 1=\log(1+N)>\log N$$

Edit :

Let $\;f(x)=\log(1+x)-x$

$$\implies f'(x)=\frac1{1+x}-1=-\frac{x}{1+x}<0\;,\;\;\text{ if }x>0\;.$$

So $f(x)$ being a decreasing function of $\;x\;$ for $\;x\geqslant0\;,$

$$f(x)<f(0)\implies\log(1+x)<x \;,\;\;\text{ if }x>0\;.$$

Angelo
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Z Ahmed
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  • This answer shows that option 4 is correct and option 3 is false. What about options 1 and 2 ? – Mera bhai Jun 12 '21 at 17:15
  • The option 1 is true. The option 2 is false for $N=1$. You can prove that the option 1 is true by using Joe’s hint which is $$\log x=\int_1^x\frac1t,dt$$ – Angelo Jun 12 '21 at 17:26
  • @Mera bhai Option 1 is correct. see my recent answer for https://math.stackexchange.com/questions/306371/simple-proof-of-showing-the-harmonic-number-h-n-theta-log-n – Z Ahmed Jun 14 '21 at 06:40
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Note that $\log x = \int_{1}^{x}\frac{1}{t} \, dt$, meaning that the question invites a geometric solution. Below is a plot of $y=\frac{1}{t}$. By comparing the red areas with the area under the curve, convince yourself that $\frac{1}{2} \leq \int_{1}^{2}\frac{1}{t} \, dt$, and $\frac{1}{3}\leq \int_{2}^{3}\frac{1}{t} \, dt$, etc. What can you infer from this?

Lower sums

Joe
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  • On summing above inequities, we get $\sum_{n = 1}^N\frac{1}{n} \leq \int_{1}^{N}\frac{1}{t} , dt = log N$ which contradict above answers. – Mera bhai Jun 12 '21 at 18:12
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    @Merabhai: That's not true. Note that the sum of the red areas is $\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{N}$, i.e. the sum starts at $\frac{1}{2}$, not $1$. – Joe Jun 12 '21 at 18:14
  • If the sum starts at $\frac{1}{2}$ then Is inequalities revert ? – Mera bhai Jun 12 '21 at 18:25
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    @Merabhai: I don't understand what you mean. What my answer shows is that $\sum_{n=2}^{N}\frac{1}{n}\leq\log N$. Add $1$ to both sides... – Joe Jun 12 '21 at 18:31
  • It means options 1,2 are also correct. But Angelo said that option 2 is false. – Mera bhai Jun 12 '21 at 18:35
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    @Merabhai: The above diagram assumes that $N\geq 2$ (there's a limit to how useful any geometric aid can be). The case $N=1$ has to be dealt with separately. Just plug in $N=1$ and see which answer is correct. – Joe Jun 12 '21 at 18:40
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    @Merabhai, I just said that the option 2 is false for $N=1$, not for $N\geqslant2$, indeed the option 2 is true for any $N\geqslant2$. – Angelo Jun 12 '21 at 19:30