- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\log(2)\)

There might be more than one way to prove it

- Thread starter ZaidAlyafey
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\log(2)\)

There might be more than one way to prove it

- Jan 31, 2012

- 253

- Jan 31, 2012

- 253

Then

$$\lim_{s \to 1} \ \eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \lim_{s \to 1 } \ (1-2^{1-s}) \zeta(s) = \lim_{s \to 1} (1-2^{1-s}) \Big( \frac{1}{s-1} + \mathcal{O}(1) \Big)$$

$$= \lim_{s \to 1} \frac{1-2^{1-s}}{s-1} = \lim_{s \to 1} \frac{2^{1-s} \log 2}{1} = \log 2 $$

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