# A proof with a mistake ( or not ) Part 2

**Number Theory**Level 3

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 \( \pi = 4 \)!

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