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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:



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What is $N$ in the first one? What is $d$ in the second one? –  Aryabhata Oct 14 '10 at 18:20
I changed \phi to \varphi, if you don't like it you can hit revert but I think that is an improvement. –  anon Oct 14 '10 at 18:23
$N$ in the first one and $d$ in the second one are positive integers. –  Nick Oct 14 '10 at 18:35
Posted also in mathoverflow.net/questions/42181/… –  Arturo Magidin Oct 14 '10 at 19:21

1 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 $\frac{1}{d!}$.

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