How is $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}= \log 2?$$
I haven't done sequences in a long time, therefore proving this seems almost impossible. How is this sum gotten. Help very much appreciated.
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2Look up the Taylor series expansion of $\ln(1+x)$. – Gregory Grant Nov 16 '15 at 15:47
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I thought about this earlier, but couldn't put it together. – Bozo Vulicevic Nov 16 '15 at 15:49
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7That, plus Abel's theorem to show that the series is still equal to the function at points on the circle of convergence where the series converges. – Robert Israel Nov 16 '15 at 15:50
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hint: $\ln 2 = \ln(1+1)$ – John Joy Nov 16 '15 at 23:59
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Hi Professor Israel, I also worked this problem out a few days ago, but using pretty much only the technique that Harry uses in his answer below. I swapped a finite integral with a finite sum (leaving the limit outside for the moment), then used dominated convergence theorem to move the limit inside of the integral. – User001 Nov 18 '15 at 23:32
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The result is just integration of the (convergent) sum 1/(1+x), over the interval [0,1], or, integration of 1/u over the interval [1,2], which gives the ln(2), as needed. Do you think that there is anything missing from this approach, i.e., do I still need Abel's theorem? The logic of my work seems fine and I think complete, but I just want to be sure -- I could have overlooked something important. Thanks @RobertIsrael. – User001 Nov 18 '15 at 23:32
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And proceeding in this way, there was no need to worry about the convergence on the circle / boundary, I think, @RobertIsrael. – User001 Nov 18 '15 at 23:36
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Yes, you can do it this way, since the partial sums are uniformly bounded. – Robert Israel Nov 19 '15 at 00:17
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\begin{align*} f(x) = \sum_{n=1}^{\infty} \frac{x^{n+1}}{n} &= x \sum_{n=1}^{\infty} \frac{x^n}{n} \\ & = x \sum_{n=1}^{\infty} \int x^{n-1} dx \\ & = x \sum_{n=0}^{\infty} \int x^n dx \\ & = x \int \left ( \sum_{n=0}^{\infty} x^n \right ) dx \\ & = x \int \left ( \frac{1}{1-x} \right ) dx \\ & = -x \log \vert 1-x \vert \\ \end{align*} \begin{align*} So, f(-1) &= -(-1) \log |1-(-1)| \\ & = \log(2) \end{align*}
