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$2 <(1+\frac{1}{n})^{n}< 3$

Is it possible to show the inequality without using binommial coefficients thus only by induction? The leftinequality can be shown using bernoulli inequality.

RM777
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  • I suspect the fact that $(1-\frac16)^{-6}$ is (barely) less than $3$ will be relevant. – Akiva Weinberger Nov 06 '18 at 18:54
  • If this must be true for all $n$ then it is false unless you change the first inequality to $\le$. – John Douma Nov 06 '18 at 19:01
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    You can find a proof by creative telescoping at page 106 here – Jack D'Aurizio Nov 06 '18 at 19:05
  • yes n>1 and I am looking for a proof that would only use tje ordering axioms and or induction. It is one of the first exc. of my book we have not introduced binommial coefficients yet or sequences or log so it wouldnt make sense using these methods in a proof. – RM777 Nov 06 '18 at 21:19

3 Answers3

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Let $a_n=\left(1+\frac{1}{n}\right)^n$. For any $n>1$ the inequality $a_n>2$ is trivial.
We have $a_{n+1}>a_n$ by AM-GM: $$\sqrt[n+1]{1\cdot a_n} = \text{GM}\left(1,1+\tfrac{1}{n},\ldots,1+\tfrac{1}{n}\right) \stackrel{\text{AM-GM}}{<} \tfrac{1}{n+1}\left[1+n\cdot\left(1+\tfrac{1}{n}\right)\right] = 1+\tfrac{1}{n+1}.$$ Additionally: $$ \frac{a_{2n}}{a_n} = \left(1+\frac{1}{4n(n+1)}\right)^n \leq \frac{1}{1-\frac{1}{4(n+1)}} = 1+\frac{1}{4n+3}$$ hence for any $N\geq 1$ we have: $$ a_N \leq a_1 \prod_{k\geq 0}\left(1+\frac{1}{4\cdot 2^k+3}\right)=\frac{16}{7}\prod_{k\geq 1}\left(1+\frac{1}{4\cdot 2^k+3}\right)\leq \frac{16}{7}\prod_{k\geq 1}\frac{1+\frac{1}{2^{k+1}}}{1+\frac{1}{2^{k+2}}}=\frac{20}{7}.$$

Jack D'Aurizio
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  • We have not introduced roots yet but I searched for the definition of telescoping in the link you provided. And I noticed that there were some similarities with a formula that was in the scnd. exc. of my book (this question deals with the third exc.) the formula which had to be proved is $ a^{n+1} - b^{n+1} = (a-b)(\sum_{i=0}^{n} a^{n-i}b^{i})$ do you think I can use this somehow to prove this inequality? – RM777 Nov 06 '18 at 23:35
  • What is AM and GM? – RM777 Nov 07 '18 at 11:41
  • @RM777: Arithmetic mean and geometric mean. – Jack D'Aurizio Nov 07 '18 at 11:43
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You have the left inequality already (Bernoulli).

The right inequality can be expressed as $$ {n} \log (1+\frac{1}{n}) < \log 3 $$ Now $\log(1+x) < x$ by concavity, which gives $1 < \log 3$ which is true.

Andreas
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I have found the answer to my problem in a different forum:

We show that $b_n > b_{n+1}$ with $b_n:= (1+\frac{1}{n})^{n+1}$

Proof:

To show: $\frac{b_n}{b_{n+1}}\geq 1$

$\frac{b_n}{b_{n+1}} = \frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}} = \frac{1}{1+\frac{1}{n}} \cdot \left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n+2} = \frac{1}{1+\frac{1}{n}} \cdot \left(1+\frac{1}{n(n+2)}\right)^{n+2}$

Exploiting the bernoulli inequality

$\frac{1}{1+\frac{1}{n}} \cdot \left(1+\frac{1}{n(n+2)}\right)^{n+2} \geq 1$

$b_5 < 3$ and $(1+\frac{1}{n})^n < b_n $

Thus $\forall n\geq 5: (1+\frac{1}{n})^n < 3$

RM777
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