5-3
)50.5-30.5)(50.5+30.5)=5-3 True
(50.25-30.25)(50.25+30.25) (50.5+30.5)=5-3 True
(51/∞-31/∞)(51/∞+31/∞)….(50.25+30.25)(50.25+30.25) (50.5+30.5)=5-3
(51/∞-31/∞)=50-30=1-1=0 Now you have gone off the rails by treating infinity as a finite number. See below.
5-3=0
????
\(\displaystyle \displaystyle n \in \mathbb Z\ and\ n > 0 \implies \left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right) = 5 - 3 = 2.\)
\(\displaystyle \displaystyle So\ \lim_{n \rightarrow \infty}\left\{\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right)\right\} = \lim_{n \rightarrow \infty} (2) = 2.\)
But you say, and truly, \(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) = 0\ and\ 0\ times\ any\ finite\ number = 0 \ne 2.\)
However, \(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\left\{\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right)\right\} = \lim_{n \rightarrow \infty}\left(5^{(1/2)^n} - 3^{(1/2)^n)}\right) * \lim_{n \rightarrow \infty} \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right).\)
And \(\displaystyle \displaystyle\lim_{n \rightarrow \infty} \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right) = \infty.\)
\(\displaystyle 0 * \infty = 2\) is true. You are not multiplying 0 times a finite quantity.
How do we know that \(\displaystyle \displaystyle\lim_{n \rightarrow \infty} \prod_{i = 1}^n\left(5^{(1/2)^i} + 3^{(1/2)^i}\right) = \infty.\)
\(\displaystyle 5^{(1/2)^i} > 1 < 3^{(1/2)^i} \implies 5^{(1/2)^i} + 3^{(1/2)^i} > 2 \implies\)
\(\displaystyle \displaystyle \prod_{i = 1}^n\left( 5^{(1/n)^i} + 3^{(1/n)^i}\right) > 2^n \implies\)
\(\displaystyle \displaystyle \lim_{n \rightarrow \infty} \prod_{i = 1}^n\left( 5^{(1/n)^i} + 3^{(1/n)^i}\right) \ge \lim_{n \rightarrow \infty}2^n = \infty.\)
