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The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.
First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$

Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$ Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$
Thanks in advance for any hint.

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    $\begingroup$ Show that $n \mapsto \left( {1 + \frac{1}{n}} \right)^n $ is strictly increasing: by the AM-GM inequality $$\left( {1 + \frac{1}{n}} \right)^n = \left( {1 + \frac{1}{n}} \right)^n \cdot 1 < \left[ {\frac{{n\left( {1 + \frac{1}{n}} \right) + 1}}{{n + 1}}} \right]^{n + 1} = \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} . $$ $\endgroup$
    – Gary
    Mar 4, 2021 at 12:22
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    $\begingroup$ "$1000$ is big enough ..." is a bit unconvincing. $(1.001)^x=2$ at around $693$. This is mere luck if you ask me $\endgroup$
    – DatBoi
    Mar 4, 2021 at 12:26
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    $\begingroup$ I think your binomial expansion approach is pretty simple and natural. You don't even need to evaluate the third term - just see that it is >0. $\endgroup$
    – Blitzer
    Mar 4, 2021 at 12:29
  • $\begingroup$ @Gary: Thank you but after that ? $\endgroup$
    – Khosrotash
    Mar 4, 2021 at 13:41
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    $\begingroup$ @Khosrotash Then $2 = a_1 < a_{1000} $. ($a_n=\left(1+\frac{1}{n}\right)^n$) $\endgroup$
    – Gary
    Mar 4, 2021 at 13:48

6 Answers 6

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According to Bernoullli's inequality, we have $(1+x)^r\ge1+xr$. for this problem consider $x=0.001$ and $r=1000$.

Note: Here $(1+x)^r=1+xr$ is not the case because $r\neq0$

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  • $\begingroup$ I wouldn't say that using only the first two terms in the binomial expansion is 'simpler' than the binomial expansion itself. $\endgroup$
    – Brady Gilg
    Mar 10, 2021 at 18:28
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The inequality is equivalent to

$$2^{\frac 1{1000}} <1+\frac 1{1000} $$

But this is true according to GM-AM:

$$2^{\frac 1{1000}} = \sqrt[1000]{1^{999}\cdot 2}\stackrel{GM-AM}{<}\frac{999+2}{1000} = 1+\frac 1{1000}$$

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  • $\begingroup$ VERY nice +1. i enjoy....this one is very simple to understand. $\endgroup$
    – Khosrotash
    Mar 4, 2021 at 19:25
  • $\begingroup$ @Khosrotash Thank you for your appreciation :-) $\endgroup$ Mar 4, 2021 at 19:26
  • $\begingroup$ I do appreciate it, but as time goes new ideas coming... I want to take tickmark to answer, but let me know if there is more solution. $\endgroup$
    – Khosrotash
    Mar 4, 2021 at 19:28
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By elementary arithmetic:

$$1.001^2=1.002001\ge1.002$$

$$1.001^4=1.004006004001\ge1.004$$ $$\cdots$$

$$1.001^{128}>1.128.$$ $$1.001^{256}>1.256.$$ $$1.001^{512}>1.512.$$

Finally,

$$1.001^{1000}>1.001^{512+256+128}>1.512\cdot1.256\cdot1.128>2.$$

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  • $\begingroup$ very nice and easy to understand for a child who never deals with calculus technics. $\endgroup$
    – Khosrotash
    Mar 4, 2021 at 13:36
  • $\begingroup$ @Khosrotash: thanks, that's what I am ;-) $\endgroup$
    – user65203
    Mar 4, 2021 at 13:44
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Let's pretend we know nothing of mathematical induction, so we can't just say $(1+x)^n\ge1+nx$ for all $n\ge1$ (and $x\ge-1$), but let's assume we can do enough algebra to know that

$$(1+x)^2=1+2x+x^2\gt1+2x$$

and

$$(1+x)^5=1+5x+10x^2+10x^3+5x^4+x^5\gt1+5x$$

for all $x\gt0$. Then we can show

$$(1+x)^{10}\gt(1+5x)^2\gt1+10x$$

from which it follows that

$$(1+x)^{100}\gt(1+10x)^{10}\gt1+100x$$

and, finally,

$$(1+x)^{1000}\gt(1+100x)^{10}\gt1+1000x$$

at which point we can plug in $x=1/1000$ to conclude

$$1.001^{1000}=\left(1+{1\over1000}\right)^{1000}\gt1+{1000\over1000}=2$$

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You can us bernoulli inequality

$(1,001)^{1000}=(\frac{1001}{1000})^{1000}=$

$(1+\frac{1}{1000})^{1000}\geq 1+\frac{1} {1000}×1000$

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$$f(x):=\left(1+\frac1x\right)^x$$ is a strictly growing function of $x$, and $f(1)=2$.


Indeed for $x>0$,

$$(\log(f(x))'=\log\left(1+\frac1x\right)-\frac1{x+1}>0$$

because

$$(\log(f(x))''=-\frac1{x(x+1)^2}<0$$ and the first derivative decreases to $0$.

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