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It was shown in here that $\left(1+\frac{1}{n}\right)^n < n$ for $n>3$. I think we can be come up with a better bound, as follows:

$$\left(1+\frac{1}{n}\right)^n \le 3-\frac{1}{n}$$ for all natural number $n$.

The result is true for all real number $\ge 1$, which can be shown using calculus. I wonder if the above result can be proved using mathematical induction?

I have tried but fail! Anyway, this question is also inspired by, and related to this question.

Edit:

I also found that $$\left(1+\frac{1}{n+k}\right)^n \le 3-\frac{k+1}{n}$$ for all natural number $k$, some large $N$ and $n > N$. This implies that $$\left(1+\frac{1}{2n}\right)^n \le 2-\frac{1}{n}.$$

And again, I can't prove any of them using Mathematical Induction.

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  • $\begingroup$ In order to use induction, I need to have $\left(1+\frac{1}{n+1}\right)^n < 3-\frac{2}{n}$ for $n\ge 2$, which again is not trivial. Equivalently, it is to prove $\left(1+\frac{1}{n}\right)^n -\left(1+\frac{1}{n+1}\right)^n >\frac{1}{n}$ for $n \ge 2$. $\endgroup$ – pipi Feb 20 '13 at 0:53
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I prove this inequality does not use induction, but I think this proof also elementary proof, because this proof does not use calculus.

$$\begin{array}{lcl} \left( 1+\frac{1}{n}\right)^n &=& 1+\binom{n}{1} \frac{1}{n}+ \sum_{k=2}^n\binom{n}{k} \frac{1}{n^k}\\ &=&2+\sum_{k=2}^n \frac{1}{k!} \frac{n(n-1)\cdots (n-k+1)}{n^k}\\ &\le& 2+\sum_{k=2}^n \frac{1}{k!} \le 2+\sum_{k=2}^n \frac{1}{k(k-1)} \\ &=& 3-\frac{1}{n} \end{array} $$

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  • 1
    $\begingroup$ Typo, 1st line, right side, $n$ should be $1/n$. $\endgroup$ – Gerry Myerson Feb 19 '13 at 4:52
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    $\begingroup$ Nice proof. +1. $\endgroup$ – user1551 Feb 19 '13 at 4:55
  • $\begingroup$ @GerryMyerson Thanks! $\endgroup$ – Hanul Jeon Feb 19 '13 at 4:56
  • $\begingroup$ @tetori, thanks for the nice proof. Anyway, I am thinking of mathematical induction... $\endgroup$ – pipi Feb 19 '13 at 5:26
  • $\begingroup$ So simple and elegant! +1 $\endgroup$ – Mark Viola Aug 26 '15 at 18:17

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