# A limit using the Euler number: $\lim_{n\rightarrow\infty}\frac{n!}{(n-i)!}\left(\frac{c}{n}\right)^{n-i}$

What is answer of this limit and how can I get it? $c$ and $i$ are constants. $$\lim_{n\rightarrow\infty}\frac{n!}{(n-i)!}\left(\frac{c}{n}\right)^{n-i}$$ I guess it will envolve some Neper/the Euler number $e$. I tried to rearrange terms of factorial and exponents in a good way but I couldn't make any conclusion so I think couldn't find the nice shape of writing this expression.

• Isn't $\;c\;$ given? Jul 15 '16 at 17:22
• The question is unclear. You should at least describe what $c$ and $i$ are. For example, what range does $i$ belong to? Jul 15 '16 at 17:26
• In Italian, the Neper number is the constant $e$. It's from John Napier. Jul 15 '16 at 17:32
• @H.W. Sorry, I only know English.
– user940
Jul 15 '16 at 17:37
• It turns out that $e$ plays no role here, since $\frac{n!}{(n-1)!}\le n^i$, and therefore our expression is $\le c^i (c/n)^{n-2i}$. Jul 15 '16 at 19:39

Hint. We assume $c,i$ are any fixed real numbers.
By setting $\displaystyle u_n=\frac{n!}{(n-i)!}\left(\frac{c}{n}\right)^{n-i}$, one gets, as $n \to \infty$, \begin{align} \frac{u_{n+1}}{u_n}&=\frac{(n+1)!}{(n+1-i)!}\left(\frac{c}{n+1}\right)^{n+1-i}\cdot \frac{(n-i)!}{n!}\left(\frac{n}{c}\right)^{n-i} \\\\&=\frac{(n+1)}{(n+1-i)}\cdot \frac{c}{n+1}\cdot \left(1+\frac1n \right)^{-n+i} \\\\& \sim \frac{c}{e(n+1)} \\\\& \to 0. \end{align} Can you take it from here?
• @H.W. bigger than $e$. Jul 15 '16 at 17:40
• @Olivier Haven't you forgotten a factor $\frac{c}{n+1}$? When using Stirling, I found the original expression $u_n$ to be equivalent to $n^i\left(\frac{c}{n}\right)^{n-i}$, which goes to $0$ (quite fast) regardless of the value of $c$. Jul 15 '16 at 17:46
• For an upper bound, one needs very little, since $n!/(n-i)!\le n^i$. Jul 15 '16 at 17:56