Calculate the sum $\sum_{n=1}^\infty \frac{2^nx^n}{n}$ I want to calculate the radius of convergence and the sum of
$$\sum_{n=1}^\infty \frac{2^nx^n}{n}$$
I have already calculed the radius of convergence using the Cauchy–Hadamard theorem (or using the ratio test), it is $|x|<\frac{1}{2}$. Now, how can I calculate the sum? For $x>\frac{1}{2}$ its easy to see that $\sum_{n=1}^\infty \frac{2^nx^n}{n}=\infty$. How do i calculate it for $x<\frac{1}{2}$?
 A: $f(x) = \sum_\limits{n=1}^{\infty}\frac {2^n x^n}{n}$
You might note that
$\frac {df}{dx} = \sum_\limits{n=1}^{\infty}2^n x^{n-1}$
Sum of a geometric series (when it converges)
$\frac {df}{dx} = \frac {2}{1-2x}$
Integrate to find $f(x)$
A: $$
\frac d {dy} \,\frac{y^n} n = y^{n-1} 
$$
$$
\frac d {dy} \sum_{n=1}^\infty \frac{y^n} n = \sum_{n=1}^\infty y^{n-1} = \frac 1 {1-y}
$$
$$
\sum_{n=1}^\infty \frac{y^n} n = \text{constant} -\log(1-y) 
$$
Since these must be equal when $y=0,$ the "constant" must be $0.$
Now let $y=2x$ and you get
$$
\sum_{n=1}^\infty \frac{2^n x^n} n = - \log(1-2x).
$$
A: This series is closely related to taylor series of logarithm
$$\ln(1+x) = x- \frac{x^2}{2}+\frac{x^3}{3}-\cdots$$
A: Also related to $\ln(\frac{1}{1-2x})$ (-:


*

*Learn a proof for convergence of Maclaurin/Taylor series when given a differentiable function.

2 . Manually prove that $\ln(1+x)=x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+....$


*substitue $1+x$ with $1-2x$ notice that $- \ln x =\ln x^{-1}$ and use 2. with the expansion of the resulting $\ln\frac{1}{1-2x}$

