Convergence and existence of constants

[evinda]

Hello! (Wave)

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)

 

[Opalg]

Hello! (Wave)

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)

That is correct so far. You could continue like this (dividing each of the factors in the numerator by one of the $n$s in the denominator):
\[\frac{(n-(m-1)) \cdots n}{m!} n^{-m} = \frac1{m!}\frac{(n-(m-1)) \cdots (n-1)\cdot n}{n^m} = \frac1{m!}\left(1-\frac{m-1}n\right) \cdots \left(1-\frac1n\right).\]
Can you finish it from there?

 

[evinda]

That is correct so far. You could continue like this (dividing each of the factors in the numerator by one of the $n$s in the denominator):
\[\frac{(n-(m-1)) \cdots n}{m!} n^{-m} = \frac1{m!}\frac{(n-(m-1)) \cdots (n-1)\cdot n}{n^m} = \frac1{m!}\left(1-\frac{m-1}n\right) \cdots \left(1-\frac1n\right).\]
Can you finish it from there?

Since $\lim_{n \to +\infty} \frac{x}{n}=0, \forall x \in [1,m-1]$, it follows that $\lim_{n \to +\infty} \binom{n}{m} n^{-m}=\frac{1}{m!}$, right?

 

[Opalg]

Since $\lim_{n \to +\infty} \frac{x}{n}=0, \forall x \in [1,m-1]$, it follows that $\lim_{n \to +\infty} \binom{n}{m} n^{-m}=\frac{1}{m!}$, right?

Yes, that's right. It's the theorem that says that the limit of a product is the product of the limits. Here, you have a fixed finite number (namely $m-1$) of terms, each of which is converging to $1$, so their product also converges to $1$.

 

[evinda]

Yes, that's right. It's the theorem that says that the limit of a product is the product of the limits. Here, you have a fixed finite number (namely $m-1$) of terms, each of which is converging to $1$, so their product also converges to $1$.

Yes! (Nod)

Since $\lim_{n \to +\infty} \binom{n}{m} n^{-m}=\frac{1}{m!}$, we have that $\forall \epsilon>0$, $\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$,

$$\left| \binom{n}{m} n^{-m}-\frac{1}{m!}\right|< \epsilon \Rightarrow \frac{1}{m!}-\epsilon< \binom{n}{m} n^{-m}<\frac{1}{m!}+\epsilon \Rightarrow \left( \frac{1}{m!}-\epsilon\right) n^m< \binom{n}{m}< \left( \frac{1}{m!}+\epsilon\right)n^m$$

So like that, we have shown that there are constants $C_1>0, C_2>0$ and a natural number $n_0$ such that $C_1 n^m \leq \binom{n}{m} \leq C_2 n^m$ for each $n \geq n_0$, right? (Smile)

 

[Opalg]

Yes! (Nod)

Since $\lim_{n \to +\infty} \binom{n}{m} n^{-m}=\frac{1}{m!}$, we have that $\forall \epsilon>0$, $\exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$,

$$\left| \binom{n}{m} n^{-m}-\frac{1}{m!}\right|< \epsilon \Rightarrow \frac{1}{m!}-\epsilon< \binom{n}{m} n^{-m}<\frac{1}{m!}+\epsilon \Rightarrow \left( \frac{1}{m!}-\epsilon\right) n^m< \binom{n}{m}< \left( \frac{1}{m!}+\epsilon\right)n^m$$

So like that, we have shown that there are constants $C_1>0, C_2>0$ and a natural number $n_0$ such that $C_1 n^m \leq \binom{n}{m} \leq C_2 n^m$ for each $n \geq n_0$, right? (Smile)

Good! :)

The only thing you have to be careful about is to choose $\epsilon < \frac1{m!}$ (otherwise $C_1$ would not be positive).

 

[evinda]

Good! :)

The only thing you have to be careful about is to choose $\epsilon < \frac1{m!}$ (otherwise $C_1$ would not be positive).

I see... Thanks a lot... (Smile)