Limit involving the totient function and combination

Question

Do you think the following limits are correct?

$\displaystyle\lim_{d\to\infty}\frac{\sum\limits_{k=1}^{d} {\varphi(N) \choose k} {d-1 \choose k-1}}{\varphi(N)^d}=0$

$\displaystyle\lim_{N\to\infty}\frac{\sum\limits_{k=1}^{d} {\varphi(N) \choose k} {d-1 \choose k-1}}{\varphi(N)^d}=c$

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/

Thanks,

Answer 1

From MathOverflow user JBL:

We have the Vandermonde identity:

$$ \sum_{k=1}^d {\varphi(N) \choose k} {d-1 \choose k-1} = {d + \varphi(N) - 1 \choose d}. $$

Thus with $N$ fixed, the numerator of your fraction is polynomial in $d$ and the result follows (with the exception of the values $N = 1, 2$).

The second result follows by the same analysis, since $\varphi(N) \to \infty$ as $N \to \infty$. In particular, the resulting constant is $\frac{1}{d!}$.