Suppose that $1< b< 2$ is a fixed constant, show that $n^b \times \prod_{k=2}^{n} (1-\frac{b}{k} ) \space \space \forall n\ge 2$ is bounded.

Question

Suppose that $1< b< 2$ is a fixed constant and $$a_n = \prod_{k=2}^{n} \left(1-\frac{b}{k} \right) \space \space \forall n\ge 2$$ Show that there are constants $0< m< M$ such that $m< n^ba_n< M$ for all $n\ge 2$

My first attempt is ...$$\left ( 1-\frac{b}{2} \right ) \left ( 1-\frac{b}{2} \right ) \left ( 1-\frac{b}{2} \right ) \dots \left ( 1-\frac{b}{2} \right ) < a_n < \left ( 1-\frac{1}{2} \right ) \left ( 1-\frac{1}{3} \right ) \left ( 1-\frac{1}{4} \right ) \dots \left ( 1-\frac{1}{n} \right ) $$ then...$$ \left ( 1-\frac{b}{2} \right ) ^{n-1}< a_n< \frac{1}{n} $$ so...$$ n^b\left ( 1-\frac{b}{2} \right ) ^{n-1}< n^ba_n< n^{b-1}$$ but it is clearly not bounded by two "constants" m and M

My second thought is that$$\prod_{k=1}^{\infty } \left ( 1+\alpha_k \right ) $$ converges if... $$\space \alpha_k > -1 ,\qquad \sum_{k=1}^{\infty } \left | \alpha_k \right | \space\text{ converges }$$ but I'm not sure if this can help here or not

Is there any suggestion please?

Answer 1

For one direction, you can use the world's most useful inequality: $1+x\leq e^{x}$. Specifically:

$$\prod_{k=2}^n (1-\frac{b}{k}) \leq e^{-b\sum_{k=2}^n \frac{1}{k}} $$

Recall $\sum_{k=2}^n \frac{1}{k} \geq \log(n)-1 $ and so

$$\prod_{k=2}^n (1-\frac{b}{k}) \leq e^{-b(\log(n)-1)} = e^{b} \frac{1}{n^b}$$

For the other direction, we have $1-x\geq e^{-x/(1-x)}$

$$\prod_{k=2}^n (1-\frac{b}{k}) \geq e^{-\sum_{k=2}^n \frac{b/k}{1-b/k}} = e^{-b \sum_{k=2}^n \frac{1}{k-b}} \geq e^{-b \left( \frac{1}{2-b}+\sum_{k=3}^{n+2} \frac{1}{k-2} \right)} \geq e^{-b(\frac{1}{2-b}+\log(n)+1)} = e^{-\frac{b(3-b)}{2-b}} \frac{1}{n^b}$$

So $ e^{-\frac{b(3-b)}{2-b}} \leq n^b \prod_{k=2}^n (1-\frac{b}{k}) \leq e^b$

Answer 2

Let for all $n$, $b_n = n^ba_n = n^b\prod_{k = 2}^n \left(1 - \frac{b}{k}\right)$. Clearly, for all $n$, $b_n > 0$ and, $$ \ln(b_n) = b\ln(n) + \sum_{k = 2}^n \ln\left(1 - \frac{b}{k}\right) $$ I let you check that for all $0 \leqslant x < 1$, $\ln(1 - x) \leqslant -x - \frac{x^2}{2}$ by studying the function $x \mapsto \ln(1 - x) + x + \frac{x^2}{2}$. Therefore, \begin{align*} \ln(b_n) & \leqslant b\ln(n) + \sum_{k = 2}^n -\frac{b}{k} - \frac{b^2}{2k^2}\\ & = b\ln(n) - bH_n + b - \frac{b^2}{2}\sum_{k = 2}^n \frac{1}{k^2}\\ & \leqslant b\ln(n) - b(\ln(n) + \gamma) + b - \frac{b^2}{2}\left(\frac{\pi^2}{6} - 1\right)\\ & = b(1 - \gamma) - \frac{b^2}{2}\left(\frac{\pi^2}{6} - 1\right) \end{align*} So $M = \exp\left(b(1 - \gamma) - \frac{b^2}{2}\left(\frac{\pi^2}{6} - 1\right)\right)$ fits.

You can also simply use $\ln(1 - x) \leqslant -x$ but you will get a higher bound (you will in fact obtain $M = e^{b(1 - \gamma)}$). Similarly, you can use for all $p$, $\ln(1 - x) \leqslant -\sum_{k = 1}^p \frac{x^p}{p}$ to get sharper bounds or even $\ln(1 - x) = -\sum_{k = 1}^{+\infty} \frac{x^p}{p}$ to get the exact limit of $b_n$.

For $m$, you have by concavity of $x \mapsto \ln(1 - x)$ that for all $0 \leqslant x \leqslant \frac{b}{2}$, $\ln(1 - x) \leqslant \frac{2x}{b}\ln\left(1 - \frac{b}{2}\right)$ and you can use the same trick to bound $\ln(b_n)$ from below.

Answer 3

Since, we have $b$ positive; ($1 < b < 2$)

$$n^b\prod_{k = 2}^{n}{1 - \dfrac{b}{k}} = n^b\cdot\dfrac{1}{1 - b}\cdot\prod_{k = 1}^{n}{\dfrac{k - b}{k}} = n^b\cdot\dfrac{(n - b)!}{(1 - b)\cdot (-b)!\cdot n!}$$

At $n = 2$, $$\displaystyle n^b\cdot\dfrac{(n - b)!}{(1 - b)\cdot (-b)!\cdot n!} = \dfrac{2^b(2 - b)}{2}$$

and

$$\lim\limits_{n \to \infty}{n^b\cdot\dfrac{(n - b)!}{(1 - b)\cdot (-b)!\cdot n!}} \to \dfrac{-1}{(b - 1)(-b)!}$$

Since the product is decreasing (one direction),

$$\dfrac{1}{\Gamma{(2 - b)}} < n^b\prod_{k = 2}^{n}{1 - \dfrac{b}{k}} < \dfrac{2^b(2 - b)}{2}$$