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Could someone give me a hint on finding the sum of all $x$ for the following power series:

$$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2n+1}}{2n+1} $$

I am pretty sure we need to compare this with $$arctan (x) = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1} $$ I'm just not sure how. Would integrating help us?

The other series: $$ \sum_{n=2}^{\infty}\frac{2^n-n}{n+1}x^n $$

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    $\begingroup$ Extract the first term of the $\arctan$ series, then multiply by $-1$. You'll get $$\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{x^{2n+1}}{2n+1}=x-\arctan x.$$ $\endgroup$ – Jack D'Aurizio Nov 8 '14 at 18:50
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This is not the full answer,this is work in progress anyway it might help author to get idea. $$\sum_{n=2}^ \infty \frac{2^nx^n}{n+1}-\frac{nx^n}{n+1}=\frac{1}{x}\sum_{n=2}^\infty\frac{2^nx^{n+1}}{n+1}-\frac{1}{x}\sum_{n=2}^\infty\frac{nx^{n+1}}{n+1}=\frac{1}{x}(-x-x^2+\sum_{n=0}^\infty\frac{2^nx^{n+1}}{n+1})-\frac{1}{x}(-\frac{x^2}{2}+\int\sum_{n=0}^\infty nx^ndx)=\frac{1}{x}(-x-x^2+\sum_{n=0}^\infty\frac{2^nx^{n+1}}{n+1})-\frac{1}{x}(-\frac{x^2}{2}+\int\frac{x}{(1-x)^2})=-1-x+\frac{1}{2x}\sum_{n=1}^\infty\frac{2^{n}x^{n}}{n}+\frac{x}{2}-\frac{1}{x}(\frac{1}{1-x}+\log(1-x)) =\frac{}{}$$

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Take the first term from the summation and then take out a factor of -1 from the remaining series.

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$$\arctan x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = x + \sum_{n=1}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = x - \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{2n+1}}{2n+1} $$

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