A proof with a mistake ( or not ) Part 2

The Leibniz formula for $\pi$ states that $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} ....... = \frac{\pi}{4}$ or $4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} ....... \right) = \pi .$ The following procedure uses the Lebniz formula to prove that $\pi=4.$ Find the incorrect step.

Step 1 - Termwise, add $4\left( \frac{1}{3} - \frac{1}{5} + \frac{1}{7} .......\right)$ to the LHS of the formula, and then subtract it at the end again, to keep the value constant at $\pi$. Doing so, we get $4\left( 1 + \frac{1}{3} - \frac{1}{3} + \frac{1}{5} - \frac{1}{5} ........\right) - 4\left( \frac{1}{3} - \frac{1}{5} + \frac{1}{7} .......\right) = \pi$

Step 2 - We can see that $\frac{1}{3} - \frac{1}{5} + \frac{1}{7} ....... = -\left(\frac{\pi}{4} - 1\right) ,$ so $4(1) -(-(\frac{\pi}{4} - 1) = \pi$

Step 3 - Simplifying,we get $(4 - 1 + \frac{\pi}{4}) = \pi$ We subtract $\frac{\pi}{4}$ from both sides to obtain $3 = \frac{3\pi}{4}$ But this means that $\pi = 4$!