Hi,


Do you think the following limits are correct? 


$\displaystyle\lim\_{d\to\infty}\frac{\sum\limits\_{k=1}^{d} {\phi(N) \choose k} {d-1 \choose k-1}}{\phi(N)^d}=0$


$\displaystyle\lim\_{N\to\infty}\frac{\sum\limits\_{k=1}^{d} {\phi(N) \choose k} {d-1 \choose k-1}}{\phi(N)^d}=c$


I plotted the equations and guessed the results according to the graphs but I could not prove them mathematically by myself. Any hints would be appreciated. Graphs are as follows:


<http://deniz.cs.utsa.edu/plots/d_vs_Eq.jpeg>


<http://deniz.cs.utsa.edu/plots/N_vs_Eq.jpeg>


Thanks,



Limit involving the totient function and combination
We have $$\sum\_{k = 1}^d \binom{\phi(N)}{k} \binom{d - 1}{k - 1} = \binom{d + \phi(N) - 1}{d}$$ (this is the Vandermonde identity). Thus, with $N$ fixed the numerator of your fraction is polynomial in $d$ and so the first result follows (except for $N = 1, 2$).


Edited to add:
Okay, and the second result follows by the same analysis, since $\phi(N) \to \infty$ as $N \to \infty$. In particular, the resulting constant is $\frac{1}{d!}$.


