Evaluate $\lim\limits_{n\to\infty}\prod\limits_{k=2}^{n}\frac{k^2+k-2}{k^2+k}$

Question

I can't find the product of a sequence. We have

$$\frac{(2+2)(2-1)}{2(2+1)}\frac{(3+2)(3-1)}{3(3+1)}...\frac{(k+2)(k-1)}{k(k+1)}$$

I am stuck with $$P=\frac{2(n+2)}{n^2(n-1)}$$ but that isn't correct. Can the squeeze theorem be used?

Answer 1

$$\begin{align}\lim_{n\to\infty}\prod_{k=2}^{n}\frac{k^2+k-2}{k^2+k}&=\lim_{n\to\infty}\prod_{k=2}^{n}\frac{(k+2)(k-1)}{k(k+1)}\\&=\lim_{n\to\infty}\frac{\prod_{k=2}^{n}(k+2)\prod_{k=2}^{n}(k-1)}{\prod_{k=2}^{n}k\prod_{k=2}^{n}(k+1)}\\&=\lim_{n\to\infty}\frac{\frac{(n+2)!}{3!}\times (n-1)!}{n!\times \frac{(n+1)!}{2!}}\\&=\lim_{n\to\infty}\frac{n+2}{3n}\\&=\lim_{n\to\infty}\frac{1+\frac 2n}{3}\\&=\frac 13\end{align}$$

Answer 2

$$\prod_{k=2}^{K}\frac{k^2+k-2}{k^2+k}=\prod_{k=2}^{K}\frac{k-1}{k}\prod_{k=2}^{K}\frac{k+2}{k+1}=\frac{1}{K}\cdot\frac{K+2}{3}$$ hence the limit equals $\color{red}{\displaystyle\frac{1}{3}}$.

Answer 3

HINT:

We have

$$\begin{align} \prod_{k=2}^n\frac{k^2+k-2}{k^2+k}&=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\frac{k-1}{k}\right)\\\\ &=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\right)\prod_{k=2}^{n} \left(\frac{k-1}{k}\right) \end{align}$$

Answer 4

Can you show $\displaystyle \prod_{k=2}^n\frac{k^2+k-2}{k^2+k}=\frac{n+2}{3n}$

The result follows easily from there.