A proof with a mistake ( or not ) Part 2

The Leibniz formula for π \pi states that 113+1517.......=π4 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} ....... = \frac{\pi}{4} or 4(113+1517.......)=π. 4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} ....... \right) = \pi . The following procedure uses the Lebniz formula to prove that π=4.\pi=4. Find the incorrect step.

Step 1 - Termwise, add 4(1315+17.......) 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 π \pi . Doing so, we get 4(1+1313+1515........)4(1315+17.......)=π 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 1315+17.......=(π41), \frac{1}{3} - \frac{1}{5} + \frac{1}{7} ....... = -\left(\frac{\pi}{4} - 1\right) , so 4(1)((π41)=π 4(1) -(-(\frac{\pi}{4} - 1) = \pi

Step 3 - Simplifying,we get (41+π4)=π (4 - 1 + \frac{\pi}{4}) = \pi We subtract π4 \frac{\pi}{4} from both sides to obtain 3=3π4 3 = \frac{3\pi}{4} But this means that π=4 \pi = 4 !

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