I know that it is converging because it is alternating series with terms getting smaller to zero. but I do not know what it converges to value
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$\begingroup$$$ S = 1- 1/2 + 1/3 - 1/4 + \ldots = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = \sum_{n=1}^{\infty}(-1)^{n+1}\int_{0}^{1}x^{n+1}dx \quad \Rightarrow $$$$ S = \int_{0}^{1}\sum_{n=1}^{\infty}(-x)^{n+1}dx = \int_{0}^{1}\frac{1}{1 + x}dx = \log 2 $$
$\endgroup$ 4 $\begingroup$$ S = 1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2n} - 2 ( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n}) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2n} - 2 \frac{1}{2} ( 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{n} ) $
Call $ H_n = 1 + \frac{1}{2} + ... \frac{1}{n} $ the harmonic series We know that $ H_n - \ln n\,$ converges to the Euler-Mascheroni constant$\gamma = 0.57...$
$S = H_{2n} - H_n = H_{2n} - \ln{2n} + H_n + \ln n + \ln 2n - \ln n = (H_{2n} - \ln 2n ) - ( H_n - \ln n ) + ln ( \frac{2n}{n} ) \to \ln( \frac{2n}{n} ) = \ln 2$
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