4
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$\displaystyle(1 + \frac{1}{n})^n < n$ for $n \gneq 3$

yes for $n = 1$ it is true

I assume it is true for $n = k$ and get

$\displaystyle(1+\frac{1}{k})^k < k$

I then go to $\displaystyle(1 +\frac{1}{k+1})^{k+1} < k+1$ and now I spend an hour doodling.

  • 0
    \left(1+\frac1n\right)^n < e = 2.7182... for n>02011-12-08

2 Answers 2

17

Consider $\left(1 + \frac{1}{n+1}\right)^{n+1} < \left(1 + \frac{1}{n}\right)^{n+1} = \left(1 + \frac{1}{n}\right)^{n}\left(1 + \frac{1}{n}\right) < n\left(1 + \frac{1}{n}\right) = n+1$ where the last inequality is due to the induction hypothesis.

9

First, the base case starts at $n=4$, not $n=1$, since it's not true for $n=1$. This is true for $n=4$ by direct calculation.

So assume $(1+\frac{1}{k})^k < k$. Then $ \left(1+\dfrac{1}{k+1}\right)^{k+1} = \left(1+\dfrac{1}{k+1}\right)^k\left(1+\dfrac{1}{k+1}\right)< \left(1+\dfrac{1}{k}\right)^k\left(1+\dfrac{1}{k}\right) $ since $1+\dfrac{1}{k+1}\lt 1+\dfrac{1}{k}$. Then apply your induction hypothesis.

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    @Tyson Sure, if you need any clarifications just ask.2011-12-08