Math Help Forum - View Single Post

ThePerfectHacker · #6 June 24th, 2007, 10:24 PM

1)This was a competition problem, though it was not posed exactly as I posted it. We will argue by induction, first for $n=3$ this is clearly true. Now say it is true for $2n-1$ and we want to show it is true for $2n+1$ people. Out of all the distances amoing the soldier consider the minimal distance. This distance is determined by two soldiers, call them $S_1$ and $S_2$ . By the problem $S_1$ must shoot toward $S_2$ and $S_2$ must shoot toward $S_1$ . Now there are two possibilities. 1)Somebody shoots at them (those two soldiers). 2)Somebody does not shoot at them. If #1 then one of those soldiers is hit twice, so it is impossible for all the soldiers to be hit because the number of people exceeds the number of bullets. And if #2then we have $2n-1$ people shooting amongst themselves. Hence it is as if those two soldiers have nothing to do with them. And hence by induction someone remains alive.

2)I saw this amazing sequence problem on a forum. He is how I would do it:
Infinite series is the answer here.
Since $(s_n)$ is convergent then $|s_n|\leq A$ for some $A>0$ . Choose $N \in \mathbb{N}$ so large that $\frac{A}{N} < 1$ . Then for all $n\geq N$ we have:
$\left( 1 + \frac{s_n}{n} \right)^n = e^{n \ln \left(1+\frac{s_n}{n} \right)} = e^{ s_n -\frac{s^2_n}{n}+...}$ .
Now,
$\lim \left( s_n - \frac{s_n^2}{n} - ... \right) = s_n - 0 + 0 ... = s_n$ (by Uniform Convergence) since $\frac{|s_n|}{n^m} \leq \frac{A}{n^m} \to 0$ .
Since $e^x$ is continous by definition we see that this converges to $e^s$ .