4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Dec 8 '11 at 11:16
  • $\begingroup$ Thanks for latexing it up. This is my first time on this site! $\endgroup$ – Tyson Dec 8 '11 at 11:24
  • 5
    $\begingroup$ Actually, much more is true: $\left(1 + \dfrac{1}{n}\right)^n < 3$. $\endgroup$ – lhf Dec 8 '11 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 Dec 8 '11 at 11:34
  • 4
    $\begingroup$ (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 Dec 8 '11 at 13:28
17
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 Dec 8 '11 at 11:25
  • $\begingroup$ @Tyson Sure, if you need any clarifications just ask. $\endgroup$ – yunone Dec 8 '11 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.