Find the interval of convergence A,of the $$\sum_{n\geq1}\left(\ln\frac{1}{2}+1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{(-1)^n}{n}\right)x^n$$ and calculate the sum for each value of $x\in A$ .
My work
$$\displaystyle{S\left( x \right) = \sum\limits_{n = 1}^\infty {\left( {\log \frac{1}{2} + 1 - \frac{1}{2} + \frac{1}{3} - .. - \frac{{{{\left( { - 1} \right)}^n}}}{n}} \right){x^n}} = \sum\limits_{n = 1}^\infty {\left( { - \log 2 + \sum\limits_{m = 1}^n {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right){x^n}} } $$ $$\displaystyle{ = \sum\limits_{n = 1}^\infty {\left( { - \sum\limits_{m = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} + \sum\limits_{m = 1}^n {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right){x^n}} = }$$
$$\displaystyle{ = - \sum\limits_{n = 1}^\infty {\left( {\sum\limits_{m = n + 1}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right){x^n}} }$$ $$\displaystyle{\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\left| {\sum\limits_{m = n + 1}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right| \cdot {{\left| x \right|}^n}}} = \left| x \right|} $$ it follows that the series converges for $\displaystyle{\left| x \right| < 1}$ and diverge for $\displaystyle{\left| x \right| > 1}$ .for $\displaystyle{x = \pm 1}$ $\displaystyle{\left| x \right| < 1} $ will be checked later(maybe(I didn't check)) $$\displaystyle{S\left( x \right) = - \sum\limits_{n = 1}^\infty {\left( {\sum\limits_{m = n + 1}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right){x^n}} = - \sum\limits_{n = 1}^\infty {\left( { - \frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n} + \sum\limits_{m = n}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right){x^n}} = }$$
$$\displaystyle{ = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n}} - \sum\limits_{n = 1}^\infty {\left( {\sum\limits_{m = n}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right){x^n}} = \log \left( {1 + x} \right) - \sum\limits_{n = 1}^\infty {{x^n}\left( {\sum\limits_{m = n}^\infty {\frac{{{{\left( { - 1} \right)}^{m - 1}}}}{m}} } \right)} }$$
For $|x|<1$ we have $$S(x) = - \sum_{n=1}^{\infty} \left( \sum_{m=n+1}^{\infty} \frac{(-1)^{m-1}}{m} \right)x^n = - \sum_{n=1}^{\infty} \left( - \frac{(-1)^{n-1}}{n} + \sum_{m=n}^{\infty} \frac{(-1)^{m-1}}{m} \right)x^n = $$
$$= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} - \sum_{n=1}^{\infty} \left( \sum_{m=n}^{\infty} \frac{(-1)^{m-1}}{m} \right)x^n = \log \left( 1+x \right) - \sum_{n=1}^{\infty} x^n\left( \sum_{m=n}^{\infty} \frac{(-1)^{m-1}}{m} \right)$$
But $\sum_{n=1}^{\infty} a_n\sum_{m=n}^{\infty} b_m = a_1\left( b_1 + b_2 + b_3 + .. \right) + a_2\left( b_2 + b_3 + b_4 + .. \right) + .. = \sum_{m=1}^{\infty} b_n\sum_{n=1}^m a_n$ (under the condition of convergence of the sums)
Then $$S(x) = - \sum_{n=1}^{\infty} \left( \sum_{m=n+1}^{\infty} \frac{(-1)^{m-1}}{m} \right)x^n = \log \left( 1+x \right) - \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m}\sum_{n=1}^m x^n = \log \left( 1+x \right) - \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m} \cdot \frac{x \cdot \left( 1-x^m \right)}{1 - x} = $$
$$= \log \left( 1+x \right) - \frac{x}{1-x}\left( \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m} - \sum_{m=1}^{\infty} \frac{(-1)^{m-1}x^m}{m} \right) = \log \left( 1+x \right) - \frac{x}{1-x}\left( \log 2 - \log \left( 1+x \right) \right) = \frac{\log \left( 1+x \right) - x \cdot \log 2}{1 - x}$$
For $x=1$ we have $S(1) = - \sum_{n=1}^{\infty} \left( \sum_{m=n+1}^{\infty} \frac{(-1)^{m-1}}{m} \right) = - \sum_{n=1}^{\infty} \left( \sum_{m=n+1}^{\infty} (-1)^{m-1}\int_0^1 x^{m-1} dx \right) =$$
$$- \int_0^1 \sum_{n=1}^{\infty} \left( \sum_{m=n+1}^{\infty} (-1)^{m-1} \right) = - \int_0^1 \sum_{n=1}^{\infty} \left( \sum_{m=1}^{\infty} (-1)^{m + n - 1} \right) = - \int_0^1 \frac{1}{1+x}\sum_{n=1}^{\infty} (-1)^{n} dx = \int_0^1 \frac{x}{1+x}\sum_{n=1}^{\infty} (-1)^{n-1} dx = $$
= $$\int_0^1 \frac{x}{(1+x)^{2}}dx = \int_0^1 \frac{ 1}{1+x}dx - \int_0^1 \frac{1}{(1+x)^{2}}dx = - \frac{1}{2} + \log 2 $$ where is also the limit $$\mathop{\lim }_{x \to 1-} \left( \frac{\log \left( 1+x \right) - x \cdot \log 2}{ 1 - x} \right)$$
For $x=-1$ we $$S(-1) = - \sum_{n=1}^{\infty} \left( \sum_{m=n+1}^{\infty} \frac{(-1)^{m-1}}{m} \right)(-1)^n = \sum_{n=1}^{\infty} \left( \sum_{m=1}^{\infty} \frac{(-1)^m}{{m + n}} \right) = \sum_{m=1}^{\infty} (-1)^m\left( \sum_{n=1}^{\infty} \frac{1}{{m + n}} \right)$$ which clearly diverges.
Finally $$S(x) = \sum_{n=1}^{\infty} \left( \log \frac{1}{2} + 1 - \frac{1}{2} + \frac{1}{3} - {\rm{ }},,{\rm{ }} - \frac{(-1)^n}{n} \right)x^n = \left\{ \begin{array}{l} \dfrac{{\log \left( 1+x \right) - x \cdot \log 2}}{ {1 - x}},\;\quad \left| x \right| < 1\\ \\ - \dfrac{1}{2} + \log 2,{\rm{ }}\quad \quad \quad \quad \;\quad \quad {\rm{ }}x = 1 \end{array} \right.$$ and diverges for $\left| x \right| > 1$
or $x=-1$