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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.

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  • $\begingroup$ I've latex-ed up your post. I wasn't entirely sure how to interpret n>/=3. I think it should be $n \gneq 3$, but I wasn't entirely sure. Is this correct? (It makes sense for the problem, etc.) $\endgroup$
    – user1729
    Commented Dec 8, 2011 at 11:16
  • $\begingroup$ Thanks for latexing it up. This is my first time on this site! $\endgroup$
    – Tyson
    Commented Dec 8, 2011 at 11:24
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    $\begingroup$ Actually, much more is true: $\left(1 + \dfrac{1}{n}\right)^n < 3$. $\endgroup$
    – lhf
    Commented Dec 8, 2011 at 11:30
  • $\begingroup$ @Tyson: just put dollar signs around your maths, and use curly brackets when you want latex to group things together. For example, a^{k+1} looks like $a^{k+1}$ while a^(k+1) looks like $a^(k+1)$. $\endgroup$
    – user1729
    Commented Dec 8, 2011 at 11:34
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    $\begingroup$ (Abstract) duplicate of Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$ (just divide both sides by $n^n$) $\endgroup$
    – Zev Chonoles
    Commented Dec 8, 2011 at 13:28

2 Answers 2

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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.

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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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    $\begingroup$ Yes I meant to start at n = 4, I don't know why I wrote n = 1 I guess I'm tired. Thank you so much for the help I'm still trying to figure it out from what you wrote. $\endgroup$
    – Tyson
    Commented Dec 8, 2011 at 11:25
  • $\begingroup$ @Tyson Sure, if you need any clarifications just ask. $\endgroup$
    – yunone
    Commented Dec 8, 2011 at 11:27

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