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The inequality $$ e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n, $$ where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.

Are there any more elementary proofs of this inequality?

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  • $\begingroup$ What's the question exactly, that this is $e^1<=3$? Or for all n? $\endgroup$ – Alec Teal Jan 1 '15 at 13:59
  • $\begingroup$ @AlecTeal The question is inequality for all $n$. (In hope that it may be an alternative way of obtaining the properties od $(e_n)$). $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 14:01
  • $\begingroup$ That says "induction" to me. $\endgroup$ – Alec Teal Jan 1 '15 at 14:02
  • $\begingroup$ @AlecTeal To me too, but without a success. $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 14:03
  • $\begingroup$ I saw $e_n$ proven to be bounded by comparing it to $f_n=\left(1+\frac{1}{n}\right)^{n+1}$, where you get $e_n<e_{n+1}<f_{n+1}<f_n$, but this only helps you for $n\geq 9$. $\endgroup$ – Arthur Jan 1 '15 at 14:03
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You can use induction.

For $n=1$, clearly $2 \leq 2$. Assume it holds for some $n-1 \in \mathbb{N}$. Then,

$$\left(1+ \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 1 + \frac{1}{n-1} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 3- \frac{1}{n-1} \right) \left(1 - \frac{1}{n} \right) = 3 - \frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)}.$$

It is left to show

$$-\frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)} \leq -\frac{1}{n}$$

or equivalently

$$ -3n+3-n+1 \leq -n+1,$$

i.e. $n \geq 1$. Since $n-1 \in \mathbb{N}$, we have $n \geq 2$ by choice, so the inequality holds.

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  • $\begingroup$ I have different inequality in the last line, but also true. +1. I am still waiting for another proofs, and then I'll choose the one for accepting. $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 14:39
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    $\begingroup$ By multiplying both sides of the second last inequality with $n(n-1)$, we get $-3(n-1) - n + 1 \leq -(n-1)$ which is equivalent to $-3n + 3 - n + 1 \leq -n + 1$, and by adding $3n-3$ to both sides, I ended up with $-n+1 \leq 2n-2$, if you're wondering. It seems simpler to just add $n-1$ to both sides, to retrieve $n \geq 1$ immediately. I changed the answer accordingly. $\endgroup$ – Huy Jan 1 '15 at 14:41
  • $\begingroup$ How did you get the factor of $1-\frac{1}{n}$ in the first line? $\endgroup$ – user84413 Aug 27 '15 at 18:20
  • $\begingroup$ I don't believe the inequality on the top line is correct for all n; for example, $(1+\frac{1}{3})^3=\frac{64}{27}>\frac{5}{3}=(3-\frac{1}{2})(1-\frac{1}{3})$. $\endgroup$ – user84413 Aug 30 '15 at 17:32
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Since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have: $$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\leq\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\tag{1}$$ hence: $$ \left(1+\frac{1}{n}\right)^n \leq \exp\left(1-\frac{1}{2n+2}\right)\leq\frac{e}{1+\frac{1}{2n+2}}=\frac{2n+2}{2n+3}e \tag{2}$$ that is stronger than $ \left(1+\frac{1}{n}\right)^n \leq 3-\frac{1}{n}$ for any $n\geq 2$.

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    $\begingroup$ Very interesting method. +1. And now I have the problem, if I should accept this one or the most elementary one. :-) $\endgroup$ – Przemysław Scherwentke Jan 3 '15 at 5:11
  • $\begingroup$ I'm sorry to bother you about this, but how did you get the last inequality in (2)? $\endgroup$ – user84413 Aug 27 '15 at 0:56
  • $\begingroup$ @user84413: it is just a consequence of $e^x\geq 1+x$. $\endgroup$ – Jack D'Aurizio Aug 27 '15 at 12:08
  • $\begingroup$ Thanks a lot! Now I understand. $\endgroup$ – user84413 Aug 27 '15 at 18:12
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We can expand using Binomial Theorem:

$\displaystyle \left(1+\frac{1}{n}\right)^n = \sum\limits_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum\limits_{k=0}^{n} \frac{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)}{k!} \le \sum\limits_{k=0}^{n} \frac{1}{k!}$

Now, $\displaystyle \sum\limits_{k=0}^{n} \frac{1}{k!} \le 2 + \sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 3 - \frac{1}{n}$

which proves the required inequality.

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    $\begingroup$ Yes, we can, but I am trying to find more elementary proof. $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 14:36
  • $\begingroup$ @PrzemysławScherwentke more elementary ? :) Then you must help me here, what kind of proof are you looking for and what all am I allowed to use so that you may call it elementary ? :-) $\endgroup$ – sciona Jan 1 '15 at 14:39
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    $\begingroup$ The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem. Are there any more elementary proofs of this inequality? I feel, that there must be a nonstandard proof. Well, one of them have already appeared. $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 14:43
  • $\begingroup$ @PrzemysławScherwentke oops ! so this is the proof using Binomial Thereom that you mentioned in OP. Thanks ! I'll try think of a different way ! :) $\endgroup$ – sciona Jan 1 '15 at 14:45

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