Limit Product Evaluation: $\displaystyle \infty$

[juantheron]

Evaluate $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\left(1+\frac{1}{4k^2-1}\right)$

 

[MarkFL]

My solution:

Spoiler

Write the limit as:

$$L=\prod_{k=1}^{\infty}\left(\frac{4k^2}{4k^2-1}\right)$$

Euler's infinite product for the sine function states:

$$\frac{\sin(x)}{x}=\prod_{k=1}^{\infty}\left(1-\left(\frac{x}{k\pi}\right)^2\right)$$

Let $$x=\frac{\pi}{2}$$:

$$\frac{\sin\left(\dfrac{\pi}{2}\right)}{\dfrac{\pi}{2}}=\prod_{k=1}^{\infty}\left(1-\left(\frac{\dfrac{\pi}{2}}{k\pi}\right)^2\right)$$

$$\frac{2}{\pi}=\prod_{k=1}^{\infty}\left(1-\left(\frac{1}{2k}\right)^2\right)=\prod_{k=1}^{\infty}\left(\frac{4k^2-1}{4k^2}\right)$$

$$\frac{\pi}{2}=\prod_{k=1}^{\infty}\left(\frac{4k^2}{4k^2-1}\right)$$

Hence:

$$L=\frac{\pi}{2}$$

This is known as Wallis' product. :D

 

[juantheron]

Thanks Markfl for Nice solution. I have solved it using Wall,s formula and sandwitch Theorem