Find a power series for $\ln(1−x)$. Use the sigma symbol below to write your final answer. Using this power series,
$$\frac { 1 }{ 1-x } =\sum _{ k=0 }^{ \infty  }{ x^ k }$$  
a) find a power series for $\ln(1 − x)$. Use the sigma symbol below to write your
final answer.  $ln(1-x)=\sum _{ k=0 }^{ \infty  }{ ? } $
b) Now write your answer from part (a) with a change of index, starting at n = 1.
(c) Using part (b), find the sum of $\sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ n2^ n }  } $
My attempt
a) 
$$f(x)=\ln(1-x),\quad f^\prime (x)=-\frac{1}{1-x},\quad f^{\prime\prime} (x)=\frac{1}{(1-x)^2}, \quad f^{\prime\prime\prime} (x)=-\frac{2}{(1-x)^3}, \\f^{\prime\prime\prime\prime} (x)=\dfrac{6}{(1-x)^4}$$
and so on.
$f(0)=0, \quad f^{\prime} (0)=-1, \quad f^{\prime\prime} (0)=1,\quad f^{\prime\prime\prime} (0)=-2,\quad f^{\prime\prime\prime\prime} (0)=6$ and so on
But, I am having difficulty to convert this into sigma notation. Please, help me with this
 A: We know that:
$${1\over 1-x} = \sum_{k=0}^\infty x^k$$
And integrating both sides:
$$\int {1\over 1-x}\,\mathrm dx = \int \sum_{k=0}^\infty x^k\, \mathrm dx$$
Then realize that:
$$-\ln |1-x| = \int {1\over 1-x} \,\mathrm dx = \int \sum_{k=0}^\infty x^k\,\mathrm dx = \sum_{k=0}^\infty {x^{k+1} \over k + 1}$$
Via term by term integration. You may notice that $\ln |1-x| \neq \ln(1-x)$, but for $|x|<1$ it models $\ln(1-x)$.
A: The integral of $\frac 1{1-x}$ is $-\log(1-x)$. So integrate with respect to $x$ termwise$$-\log(1-x)=\int\sum\limits_{k\geq0}x^k=\sum\limits_{k\geq0}\frac {x^{k+1}}{k+1}$$which converges when $|x|<1$. Can you complete the rest?
A: A power series, centered at $0$ (called a Maclaurin Series) is a series representation of a function given by
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$
Note that your $f^{(n)}(0)=(-1)^{n}(n-1)!$, which is what you calculated in the question.  $$-1=-0!,$$ $$1=1!,$$ $$-2=-2!,$$ $$6=3!,$$ $$...$$
Thus you get 
$$f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(n-1)!}{n!}x^n=\sum_{n=0}^\infty{\frac{(-1)^n}{n}x^n}$$
But now note that your first term yields $0$ in the denominator.  Thus
$$f(x)=\sum_{n=1}^\infty{\frac{(-1)^n}{n}x^n}$$
To finish the problem, note that if $x=-\frac{1}{2}$, 
$$f\left(-\frac{1}{2}\right)=\sum_{n=1}^\infty{\frac{(-1)^n}{n}\left(-\frac{1}{2}\right)^n}=\sum_{n=1}^\infty\frac{(-1)^{2n}}{n2^n}=\sum_{n=1}^\infty{\frac{1}{n2^n}}$$
and since $f(x)=\ln{(1-x)}$,
$$f\left(-\frac{1}{2}\right)=\ln{\left(\frac{3}{2}\right)}=\ln{3}-\ln{2}$$
A: $f'(x)$ is minus the left-hand side of your given series.  Can you negate and integrate the right-hand side to recover $f$?
A: If you look carefully you will notice a pattern.$$f^{(n)}(x)=(-1)^n\frac{(n-1)!}{(1-x)^n}$$Plugging in $x=0$ will give you your answer which can be used in the summation notation.
A: Since the derivative of $x\mapsto \log(1-x)$ is $-\frac{1}{1-x} = -\sum_{n=0}^\infty x^n$ for $\vert x\vert < 1$.
We may use dominated convergence as 
$$\log(1-t) = \int_0^t \frac{dx}{x-1}  = -\int_0^t \sum_{n=0}^\infty x^n dx  \;\;\text{ for }\;\;\vert t \vert < 1$$
and the series $\sum \int_0^t \vert x^n\vert dx$ converges for any real number $t$ between $-1$ and $1$, so we are able to permute $\sum$ and $\int$ symbols... then
$$\log(1-t) = -\sum_{n=0}^\infty \int_0^t x^n dx = -\sum_{n=1}^\infty \frac{t^n}{n}  \;\;\text{ for }\;\;\vert t \vert < 1$$
