So I have started by finding that $\sum_{n=1}^{\infty}\frac{x^n}{n}$ has a radius of convergence of 1. I would think that I need to use this to solve it by differentiating it and substituting in some value of x to make it my formula. I can't seem to find a solution.
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1Integrate the series $\sum_{i=0}^\infty x^n$ term by term and then shift the index. – Mustafa Said May 13 '20 at 20:12
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the sum is in $i$ or in $n$? – Riccardo.Alestra May 13 '20 at 20:13
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My mistake- it is for n @Riccardo.Alestra – strugglingstudent May 13 '20 at 20:14
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The Taylor Series for $$\log(1-x)= -\sum_{n=1}^{\infty} \frac{x^n}{n} $$
Set $x=\frac 12$ to get $$-\log(1-\frac 12) = \sum_{n=1}^{\infty} \frac{1}{n2^n} = \log 2$$
Vishu
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