What is the function represented by the power series What is the function represented by the below power series?
$$\sum_{k=1}^{\infty} \frac{x^k}{k}$$
I know that the function will be derived from the summation of $k=0$ to $\infty$ of $x^k$ but I don't know where to go from there. Thanks!
 A: Let $f(x) = \ln(1-x)+\sum_{k=1}^\infty {x^k \over k}$, for $|x| <1$. We have
$f'(x) = -{1 \over 1-x} + \sum_{k=1}^\infty x^{k-1} = 0$.
Since $f(0) = 0$, we see $f(x) = 0$ for $|x| <1$. Hence
$\sum_{k=1}^\infty {x^k \over k} = - \ln(1-x)$ for $|x|<1$.
A: Let $\displaystyle f(x) = \sum_{k \ge 1}\frac{x^k}{k}$ then $\displaystyle f'(x) = \sum_{k \ge 1}\frac{kx^{k-1}}{k} = \sum_{k \ge 1}x^{k-1} = \frac{1}{x}\sum_{k \ge 1}x^{k} = \frac{1}{(1-x)}$
And therefore $\displaystyle f(x)= \int \frac{1}{1-x}\;{dx} = -\log(1-x)+C$. Letting $x = 0$ we see that $C = 0$ therefore the function is  $-\log(1-x).$ 
A: Hint. Observe that $\frac{d}{dx}\left(\sum_{k=1}^{\infty}\frac{x^k}{k}\right)=\sum_{k=1}^{\infty}x^{k-1}=\frac{1}{1-x}$, if $|x|<1$
A: Since $|x|<1,$ then
$$\sum\limits_{k=1}^\infty\dfrac{x^k}k
 = \sum\limits_{k=0}^\infty\dfrac{x^{k+1}}{k+1} = \sum\limits_{k=0}^\infty\int\limits_0^x x^kdx =\int\limits_0^x\left(\sum\limits_{k=0}^\infty x^k\right)dx = \int\limits_0^x\dfrac{dx}{1-x} = -\ln(1-x)\bigg|_0^x,$$
$$\sum\limits_{k=1}^\infty\dfrac{x^k}k = -\ln(1-x).$$
On the other hand,
$$\sum\limits_{k=1}^\infty\dfrac{x^k}k = \operatorname{Li}_1(x) = -\ln(1-x),$$
where 
$$\operatorname{Li}_n(x) = \sum\limits_{k=1}^\infty\dfrac{x^k}{k^n}$$
is a polylogarithmic function.
