Proving $\frac1{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a<\binom{a(b+1)}a<\left(\frac{(b+1)^{b+1}}{b^b}\right)^a$

Question

Let $a\in\mathbb N$, and $b\in\mathbb R, b\geq 1$. How to prove that $$\frac{1}{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a<\binom{a(b+1)}{a}<\left(\frac{(b+1)^{b+1}}{b^b}\right)^{a}?$$

Answer 1

$\def\e{\mathrm{e}}$Lemma 1: If $$f(x) = \left( 1 + \dfrac{1}{x} \right)^x,\ g(x) = \left( 1 + \dfrac{1}{x} \right)^{x + 1}, \quad x \geqslant 1$$ then $f$ is strictly increasing but $g$ is strictly decreasing, and$$ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} g(x) = \e. $$ (This is a well-known result.)

Lemma 2: For $m\in \mathbb{N}_+$, $n \in \mathbb{R}$, $n \geqslant 1$,$$ \binom{m + n}{m} < \frac{(m + n)^{m + n}}{m^m n^n}. \tag{2.1} $$

Proof of lemma: (2.1) will be proved by induction on $m$. For $m = 1$,$$ (1) \Longleftrightarrow n + 1 < \frac{(n + 1)^{n + 1}}{n^n} \Longleftrightarrow n^n < (n + 1)^n. $$ Suppose (2.1) holds for $m$. To prove (2.1) for $m + 1$, it suffices to prove that$$ \left. \binom{m + n + 1}{m + 1} \middle/ \binom{m + n}{m} \right. \leqslant \left. \frac{(m + n + 1)^{m + n + 1}}{(m + 1)^{m + 1} n^n} \middle/ \frac{(m + n)^{m + n}}{m^m n^n} \right.. \tag{2.2} $$\begin{align*} (2.2) &\Longleftrightarrow \frac{m + n + 1}{m + 1} \leqslant \frac{m^m (m + n + 1)^{m + n + 1}}{(m + 1)^{m + 1} (m + n)^{m + n}}\\ &\Longleftrightarrow \frac{(m + 1)^m}{m^m} \leqslant \frac{(m + n + 1)^{m + n}}{(m + n)^{m + n}}\\ &\Longleftrightarrow \left( 1 + \frac{1}{m} \right)^m \leqslant \left( 1 + \frac{1}{m + n} \right)^{m + n}, \end{align*} where the last inequality holds by Lemma 1. End of induction.

Lemma 3: For $m\in \mathbb{N}_+$, $n \in \mathbb{R}$, $n \geqslant 1$,$$ \binom{m + n}{m} > \frac{1}{\e m} \frac{(m + n)^{m + n}}{m^m n^n}. \tag{3.1} $$

Proof of lemma: Again, (3.1) will be proved by induction on $m$. For $m = 1$,$$ (3.1) \Longleftrightarrow n + 1 > \frac{1}{\e}·\frac{(n + 1)^{n + 1}}{n^n} \Longleftrightarrow \left( 1 + \frac{1}{n} \right)^n < \e, $$ where the last inequality is true by Lemma 1.

Suppose (3.1) holds for $m$. To prove (3.1) for $m + 1$, it suffices to prove that$$ \left. \binom{m + n + 1}{m + 1} \middle/ \binom{m + n}{m} \right. \geqslant \left. \frac{(m + n + 1)^{m + n + 1}}{(m + 1)^{m + 2} n^n} \middle/ \frac{(m + n)^{m + n}}{m^{m + 1} n^n} \right.. \tag{3.2} $$\begin{align*} (3.2) &\Longleftrightarrow \frac{m + n + 1}{m + 1} \geqslant \frac{m^{m + 1} (m + n + 1)^{m + n + 1}}{(m + 1)^{m + 2} (m + n)^{m + n}}\\ &\Longleftrightarrow \frac{(m + 1)^{m + 1}}{m^{m + 1}} \geqslant \frac{(m + n + 1)^{m + n}}{(m + n)^{m + n}}\\ &\Longleftrightarrow \left( 1 + \frac{1}{m} \right)^{m + 1} \geqslant \left( 1 + \frac{1}{m + n} \right)^{m + n}, \end{align*} where the last inequality holds because by Lemma 1,$$ \left( 1 + \frac{1}{m} \right)^{m + 1} \geqslant \e \geqslant \left( 1 + \frac{1}{m + n} \right)^{m + n}. $$ End of induction.

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Now for $a \in \mathbb{N}_+$, $b \geqslant 1$, take $(m, n) = (a, ab)$ in Lemma 2, then$$ \binom{a(b + 1)}{a} < \frac{(ab + a)^{ab + a}}{a^a (ab)^{ab}} = \frac{a^{ab + a} (b + 1)^{ab + a}}{a^{ab + a} b^{ab}} = \left( \frac{(b + 1)^{b + 1}}{b^b} \right)^a. $$

Take $(m, n) = (a, ab)$ in Lemma 3, then$$ \binom{a(b + 1)}{a} > \frac{1}{\e a} \frac{(ab + a)^{ab + a}}{a^a (ab)^{ab}} = \frac{1}{\e a} \left( \frac{(b + 1)^{b + 1}}{b^b} \right)^a \geqslant \frac{1}{4ab} \left( \frac{(b + 1)^{b + 1}}{b^b} \right)^a. $$

Answer 2

In this question of mine (Show that $r_k^n/n \le \binom{kn}{n} < r_k^n$ where $r_k = \dfrac{k^k}{(k-1)^{k-1}}$) I show both that for $n \ge 2$, $\dfrac{r_k^n}{n+1} \le \binom{kn}{n} < r_k^n$ where $r_k = \frac{k^k}{(k-1)^{k-1}}$ and $\binom{kn}{n} \approx \sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k^k} {(k-1)^{k-1}}\right)^{n} $.