proof

[azreal]

hi,
proof by calculation, that
\(\displaystyle \frac{1}{m^k} \left(\begin{array}{cc}m\\k\end{array}\right) \) \(\displaystyle < \) \(\displaystyle \frac{1}{n^k} \left(\begin{array}{cc}n\\k\end{array}\right) \) \(\displaystyle \leq \) \(\displaystyle \frac{1}{k!} \) \(\displaystyle \leq \) \(\displaystyle \frac{1}{2^{k-1}} \)
for \(\displaystyle k, m, n, \in N \) and \(\displaystyle m < n \) and \(\displaystyle 2\leq k\leq n\)

If already tried to multiplicate with the denominator \(\displaystyle \frac{1}{m^k} \) \(\displaystyle \frac{1}{n^k} \) \(\displaystyle \frac{1}{2^{k-1}} \), also using the definition of the binomial coefficient, but it just made it even worse...

 

[azreal]

nobody?

 

[daon2]

Looks like you'll need to show

\(\displaystyle \displaystyle \dfrac{m!}{m^k}\cdot \dfrac{1}{(m-k)!}\cdot\dfrac{1}{k!}<\dfrac{n!}{n^k}\cdot \dfrac{1}{(n-k)!}\cdot\dfrac{1}{k!}\)

With a little manipulation, and letting \(\displaystyle n=m+d\) you obtain

\(\displaystyle \displaystyle \dfrac{(m+d)^k}{m^k}\dfrac{m!}{(m+d)!}\cdot \dfrac{(m-k+d)!}{(m-k)!}<1\)

If you're still confused, try letting d=1 (i.e. n=m+1) first.

 

[azreal]

Thanks so far, daon2, i proofed
\(\displaystyle \frac{1}{n^k} \left(\begin{array}{cc}n\\k\end{array}\right) \) \(\displaystyle \leq \) \(\displaystyle \frac{1}{k!} \) \(\displaystyle \leq \) \(\displaystyle \frac{1}{2^{k-1}} \)
till now separately.


\(\displaystyle \displaystyle \dfrac{(m+d)^k}{m^k}\dfrac{m!}{(m+d)!}\cdot \dfrac{(m-k+d)!}{(m-k)!}<1\)

I understand your idea and your transformation, but i can't see whythe whole term must bei < 1,

\(\displaystyle \displaystyle \dfrac{(m+d)^k}{m^k}\) and \(\displaystyle \dfrac{(m-k+d)!}{(m-k)!}<1\) are bigger then 1 (because of +d)

and \(\displaystyle \dfrac{m!}{(m+d)!}\) is less then 1