Calculate the sum $\sum_{n=1}^\infty \frac{2^nx^n}{n}$

Question

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 $x<\frac{1}{2}$?

Answer 1

$f(x) = \sum_\limits{n=1}^{\infty}\frac {2^n x^n}{n}$

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 $f(x)$

Answer 2

This series is closely related to taylor series of logarithm

$$\ln(1+x) = x- \frac{x^2}{2}+\frac{x^3}{3}-\cdots$$

Answer 3

Also related to $\ln(\frac{1}{1-2x})$ (-:

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

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

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

Answer 4

$$ \frac d {dy} \,\frac{y^n} n = y^{n-1} $$ $$ \frac d {dy} \sum_{n=1}^\infty \frac{y^n} n = \sum_{n=1}^\infty y^{n-1} = \frac 1 {1-y} $$ $$ \sum_{n=1}^\infty \frac{y^n} n = \text{constant} -\log(1-y) $$ Since these must be equal when $y=0,$ the "constant" must be $0.$

Now let $y=2x$ and you get $$ \sum_{n=1}^\infty \frac{2^n x^n} n = - \log(1-2x). $$