Is $\log{2} = 0$ ?

Here is a proof that $\log{2}$ = 0. This is not true but can you spot the wrong step in it.

$\displaystyle \log(1 + x) = \sum_{i = 1}^{\infty}{\dfrac{(-1)^k x^k}{k}} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \dots \dots \\ \text{Putting x = 1} \\ \log{2} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots \dots \\ \log{2} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots - 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} + \dots \right) \\ \log{2} = \zeta(1) - \frac{2}{2}\left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots \right) \\ \log{2} = \zeta(1) - \zeta(1) \\ \log{2} = \boxed{0} \\ \textbf{Hence Proved}$

#Calculus #Series #RiemannZetaFunction #Proofs #Paradox

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Comments

$\zeta(1) = \infty$

$\displaystyle \zeta(1) - \zeta(1) = \infty - \infty \neq 0$

Krishna Sharma - 6 years, 2 months ago

Thanks !