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Alternating sum

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
It might be well-known for you that

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

There might be more than one way to prove it :)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
The typical approach is to expand $\log(1+z)$ in a Taylor series about $z=0$ and then apply Abel's theorem. But that's not particularly interesting.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$$ \eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{s}} = (1-2^{1-s}) \sum_{n=1}^{\infty} \frac{1}{n^{s}} = (1-2^{1-s}) \zeta(s) $$

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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