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

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

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TopNewest\(\zeta(1) = \infty \)

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

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Thanks !

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Yeah, absolutely right.

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\( \zeta(1) \) does not converge. So the statement \( \zeta (1) - \zeta (1) =0 \) is like saying \( \infty - \infty = 0\) which is wrong because it is indeterminate.

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Thanks !

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Watch the limits in which the Taylor series converges. We are given that \( | x | < 1 \), and hence we cannot apply this to the case where \( x = 1 \) or \( x = -1 \), to find the value of \( \log 2 \) or \( \log 0 \).

Similarly, we cannot apply the Geometric Progression sum of \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \) to conclude that \( \frac{1}{0} = 1 + 1 + 1 + 1 + \ldots \) or that \( \frac{1}{2} = 1 - 1 + 1 - 1 + \ldots \), because \( x = 1, -1 \) are out of the range in which the formula applies.

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Thanks a Lot

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@Sandeep Bhardwaj @Calvin Lin @Ronak Agarwal @Mvs Saketh @Azhaghu Roopesh M

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Rajdeep did u understand all this??? It was really out of my mind!!

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I came up with this proof.

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@Manish Dash you are a total genius. In Open proof contest can I send snapshots or a word file.

Yep! I totally agree withLog in to reply

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Alternatively, this works as a proof that the harmonic series diverges.

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