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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5$\ln 2$ $\hspace{0in}$ – pancini Nov 28 '19 at 03:04
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4It is called the alternating harmonic series. See the Taylor series expansion of the natural logarithm about $x=1$. – hardmath Nov 28 '19 at 03:07
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$$ S = 1- 1/2 + 1/3 - 1/4 + \ldots = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = \sum_{n=1}^{\infty}(-1)^{n+1}\int_{0}^{1}x^{n+1}dx \quad \Rightarrow $$ $$ S = \int_{0}^{1}\sum_{n=1}^{\infty}(-x)^{n+1}dx = \int_{0}^{1}\frac{1}{1 + x}dx = \log 2 $$
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1Why did you multiply by $\int_{0}^{1}x^{n+1}dx=\frac{1}{n+2}$ instead of $\int_{0}^{1}x^{n-1}dx=\frac{1}{n}$? – Axion004 Nov 28 '19 at 03:35
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1Indeed, $\sum_{n=1}^\infty (-x)^{n+1} = \frac{x^2}{1+x}$. I believe it should instead be $\int_0^1 x^{n-1}\ \mathsf dx$ to obtain the series $\sum_{n=1}^\infty (-x)^{n-1} = \frac{1}{1+x}$. – Math1000 Nov 28 '19 at 03:39
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Can you explain how $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ became $\sum_{n=1}^\infty (-1)^{n+1} \int_{0}^{1} x^{n+1} dx$ ? – Robert Melikyan Aug 16 '22 at 13:03
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$ 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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