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Dec 8, 2011 at 20:21 history tweeted twitter.com/#!/StackMath/status/144874367103414274
Dec 8, 2011 at 20:16 comment added Aaron McDaid $\left(1+\frac1n\right)^n < e = 2.7182...$ for $n>0$
Dec 8, 2011 at 19:49 comment added user17522 Now let n go to infinity.
Dec 8, 2011 at 14:52 history edited J. M. ain't a mathematician
edited tags
S Dec 8, 2011 at 13:33 history suggested Paul CC BY-SA 3.0
add \displaystyle
Dec 8, 2011 at 13:32 review Suggested edits
S Dec 8, 2011 at 13:33
Dec 8, 2011 at 13:28 comment added Zev Chonoles (Abstract) duplicate of Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$ (just divide both sides by $n^n$)
Dec 8, 2011 at 13:28 history edited Jonas Meyer CC BY-SA 3.0
edited title
Dec 8, 2011 at 13:23 history edited user940 CC BY-SA 3.0
typo
Dec 8, 2011 at 11:37 comment added Tyson Thanks let me try $a^{k+1}$
Dec 8, 2011 at 11:34 comment added user1729 @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 $a^(k+1)$.
Dec 8, 2011 at 11:30 comment added lhf Actually, much more is true: $\left(1 + \dfrac{1}{n}\right)^n < 3$.
Dec 8, 2011 at 11:24 comment added Tyson Thanks for latexing it up. This is my first time on this site!
Dec 8, 2011 at 11:17 answer added yunone timeline score: 9
Dec 8, 2011 at 11:16 comment added user1729 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.)
Dec 8, 2011 at 11:16 answer added EuYu timeline score: 17
S Dec 8, 2011 at 11:14 history suggested user1729 CC BY-SA 3.0
latex-ing it up
Dec 8, 2011 at 11:13 review Suggested edits
S Dec 8, 2011 at 11:14
Dec 8, 2011 at 11:07 history asked Tyson CC BY-SA 3.0