Is how I evaluated the following product correct $\prod_{n=1}^{20}(1+\frac{2n+1}{n^2})$?

Question

The product is $$\prod_{n=1}^{20}\left(1+\frac{2n+1}{n^2}\right)$$. I can rewrite this product as $$\prod_{n=1}^{20}\left(\frac{n^2+2n+1}{n^2}\right)$$ which can be further simplified to $$\prod_{n=1}^{20}\frac{(n+1)^2}{n^2}$$. Here is the odd step I am going to do. I am going to rewrite this product as $$(\prod_{n=1}^{20}\frac{(n+1)}{n})^2$$. Now, I evaluate the product in the base $$\prod_{n=1}^{20}\frac{(n+1)}{n}$$. Writing out all the terms, I get $\frac{\color{red}{2}}{1} * \frac {\color{blue}{3}}{\color{red}{2}} * \frac {\color{red}{4}}{\color{blue}{3}} ... \frac{\color{blue}{19}}{\color{red}{18}} * \frac{\color{red}{20}}{\color{blue}{19}} * \frac {{21}}{\color{red}{20}}$. Terms of same color near each other cancel, black terms do not, I hope my use of blue and red helps anyone with colorblindness. Thus, we see that only $21 \over 1$ remains, making this product $21$. Now, we square it to get the original product of $441$. So, was it valid to write the product of a square as the square of a product?

Answer 1

Absolutely. You are using an expanded form of $(ab)^2=a^2b^2$, which relies on the fact that multiplication is commutative. Your telescoping works without pulling the square out, too. All the terms in the product are squared compared to what you did and the squares cancel nicely.