- #1

evinda

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Let $m$ be a natural number. I want to check the sequence $\left( \binom{n}{m} n^{-m}\right)$ as for the convergence and I want to show that there exist constants $C_1>0, C_2>0$ (independent of $n$) and a positive integer $n_0$ such that $C_1 n^m \leq \binom{n}{m} \leq C_2 n^m$ for each $n \geq n_0$.

We have that $\binom{n}{m} n^{-m}=\frac{n!}{m!(n-m)!} n^{-m}=\frac{1 \cdots (n-m) \cdot (n-(m-1)) \cdots n}{m!(n-m)!} n^{-m}=\frac{(n-m)! (n-(m-1) \cdots n)}{m!(n-m)!} n^{-m}=\frac{(n-(m-1)) \cdots n}{m!} n^{-m}$.

Is it right so far? If so, how can we find the limit of the last term? (Thinking)