1
$\begingroup$

I have to prove that the product $$\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$$ is greater than $\displaystyle\frac{2}{3}$.

I've tried to write $k^{3}-1$ as $(k-1)(k^{2}+k+1)$ or another ways but I couldn't finish it.

$\endgroup$
2
  • 2
    $\begingroup$ You have factored the numerator. The denominator likewise factors as $(k+1)(k^2-k+1)$. $\endgroup$ Jun 22, 2020 at 19:10
  • 1
    $\begingroup$ The infinite variation appears as problem 5.3.7.(b) in Larson's Problem Solving Through Problems, though the author of the book makes the mistake of starting the index at $1$ instead of $2,$ which would make the product $0$. amazon.com/Problem-Solving-Through-Problems-Problem-Mathematics/… $\endgroup$
    – Favst
    Jun 22, 2020 at 19:26

4 Answers 4

8
$\begingroup$

$\begin{array}\\ f(n) &=\prod_{k=2}^{n}\dfrac{k^{3}-1}{k^{3}+1}\\ &=\dfrac{\prod_{k=2}^{n}(k^{3}-1)}{\prod_{k=2}^{n}(k^{3}+1)}\\ &=\dfrac{\prod_{k=2}^{n}(k-1)(k^2+k+1)}{\prod_{k=2}^{n}(k+1)(k^2-k+1)}\\ &=\dfrac{\prod_{k=1}^{n-1}k}{\prod_{k=3}^{n+1}k}\dfrac{\prod_{k=3}^{n+1}((k-1)^2+(k-1)+1)}{\prod_{k=2}^{n}(k^2-k+1)} \qquad\text{(this is the only clever step)}\\ &=\dfrac{2}{n(n+1)}\dfrac{\prod_{k=3}^{n+1}(k^2-2k+1+k-1+1)}{\prod_{k=2}^{n}(k^2-k+1)}\\ &=\dfrac{2}{n(n+1)}\dfrac{\prod_{k=3}^{n+1}(k^2-k+1)}{\prod_{k=2}^{n}(k^2-k+1)}\\ &=\dfrac{2}{n(n+1)}\dfrac{(n+1)^2-(n+1)+1)}{2^2-2+1}\\ &=\dfrac{2}{n(n+1)}\dfrac{n^2+2n+1-n-1+1}{3}\\ &=\dfrac23\dfrac{n^2+n+1}{n^2+n}\\ &=\dfrac23(1+\dfrac1{n^2+n})\\ &\gt \dfrac23 \qquad\text{for all } n\\ \end{array} $

$\endgroup$
3
$\begingroup$

Use : $k^2 - k + 1 = (k-1)^2 + (k-1) + 1$ to cancel out some more terms.

$\endgroup$
3
$\begingroup$

Hint

$$k^3+1=(k+1)(k^2-k+1)$$

Hint 2:

$$(k+1)^2-(k+1)+1=k^2+k+1$$

$\endgroup$
1
$\begingroup$

$$\begin{align} \prod_{k=2}^{99} \frac{k^3-1}{k^3+1} &= \prod_{k=2}^{99} \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\ &= \Biggl(\prod_{k=2}^{99} \frac{k^2+k+1}{k+1}\Biggl) \Biggl(\prod_{k=2}^{99} \frac{k-1}{k^2-k+1} \Biggl) \space \space \text{(Separability of Multiple Products)} \\ &= \frac{\prod_{k=2}^{99} \frac{k^2+k+1}{k+1}}{\prod_{k=2}^{99} \frac{k^2-k+1}{k-1}} \\ &= \frac{\prod_{k=2}^{99} \frac{(k+1)^2-(k+1)+1}{k+1}}{\prod_{k=2}^{99} \frac{(k-1)^2+(k-1)+1}{k-1}} \\ &= \frac{\prod_{k=2}^{99} (k+1)-1+\frac{1}{k+1}}{\prod_{k=2}^{99} (k-1)+1+\frac{1}{k-1}} \\ &= \frac{\prod_{n=3}^{100} n-1+\frac{1}{n}}{\prod_{n=1}^{98} n+1+\frac{1}{n}} \\ &= \Biggl(\prod_{k=3}^{98} \frac{(n+\frac{1}{n})-1}{(n+\frac{1}{n})+1}\Biggl) \Biggl(\frac{\prod_{n=99}^{100} n+\frac{1}{n}-1}{\prod_{k=1}^{2} n+\frac{1}{n}+1}\Biggl) \\ &= \Biggl(\prod_{k=3}^{98} 1-\frac{2}{n+1+\frac{1}{n}}\Biggl) \Biggl(\frac{\prod_{n=99}^{100} n+\frac{1}{n}-1}{\prod_{k=1}^{2} n+\frac{1}{n}+1}\Biggl) \\ \end{align}$$ At this stage, we can make a numeric calculation:

$$\Biggl(\prod_{k=3}^{98} 1-\frac{2}{n+1+\frac{1}{n}}\Biggl)=\frac{7}{13} \cdot \frac{13}{21} \cdot \frac{21}{31} \cdots \frac{98 \cdot 99+1-2\cdot 98}{98 \cdot 99+1}=\frac{7}{98 \cdot 99 +1}$$ $$\Biggl(\frac{\prod_{n=99}^{100} n+\frac{1}{n}-1}{\prod_{k=1}^{2} n+\frac{1}{n}+1}\Biggl)=\frac{2}{3} \cdot \frac{(98 \cdot 99+1)\cdot(99 \cdot 100+1)}{7 \cdot 99 \cdot 100}$$

The resultant product is: $$\frac{2}{3} \cdot \frac{(98\cdot 99+1) \cdot (99 \cdot 100+1)}{7 \cdot 99 \cdot 100} \cdot \frac{7}{98 \cdot 99+1}=\frac{2}{3} \cdot \frac{9901}{9900} \gt \frac{2}{3}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .