Is the below alternate proof that $ \pi$ is irrational.
$$ \displaystyle{ \pi = 4 - \dfrac 43 + \dfrac 45 -\dfrac 47 + O_n \\ \text{ First assume that pi is rational and is equal to } \dfrac ab \\ \dfrac ab = 4 - \dfrac 43 + \dfrac 45 - \dfrac 47 + O_n \\ \text{ Now, assume } \Gamma (n) = \prod \limits_{c=4}^n 2c+1 \\ So, a. = 4b - \dfrac 43 b + \dfrac 45b - \dfrac 47b + O_n b \\ \Gamma a = 4b\Gamma - \dfrac 43b\Gamma + \dfrac 45 b\Gamma - \dfrac 47 b \Gamma +O_n b \Gamma \\ \text{ LHS is an integer and RHS isn't, this arises due to our wrong assumption that pi is rational, thus, pi is irrational } \blacksquare \\ \text{We can take n sufficiently large to be close enough to pi.} }$$
What's invalid in this proof here?
Although the fractions in $O_nb\Gamma$ are not integers, it might be that after you add them all up, they converge on a whole number. For example, $$\frac32-\frac34+\frac38-\frac3{16}-\frac3{32}+\cdots=1$$