Sum the series of $\sum\limits_{n=1}^{\infty} \frac{1}{2^n n} $

Question

Exactly sum the series $\sum\limits_{n=1}^{\infty} \frac{1}{2^n n} $

I understand that a power series is needed. However, I am unsure which one. I thought of using $\sum\limits_{n=1}^{\infty} \frac{1}{2^n n} x^n $ but im not sure if this is correct due to still leaving the n on the denominator - in an example I have seen with an n on the numerator of the series this disappeared for the power series. I do not understand that.

Then when finding a series I think I should find the derivative (if radius of convergence is bigger than 0), but overall am not totally sure how to progress with this problem.

Answer 1

Define

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

on the interval $x\in (-2,2)$. To solve your question, we want the value of $f(1)$. Note that this is a very well defined series on the interval $[0,1]$. We'll skip the details, but on this closed interval we can differentiate term by term. Doing just that gives

$$f'(x)=\sum_{n=1}^\infty \frac{n}{2^nn}x^{n-1}=\sum_{n=1}^\infty \frac{1}{2^n}x^{n-1}=\frac{1}{2}\sum_{n=0}^\infty \left(\frac{x}{2}\right)^n$$

This is an easy geometric series

$$=\frac{1}{2}\frac{1}{1-x/2}=\frac{1}{2-x}$$

Then

$$f(1)=f(1)-f(0)+f(0)=\int_0^1 f'(t)dt+f(0)=\int_0^1 \frac{1}{2-t}dt+0=\ln(2)$$

Answer 2

Let $ n\in\mathbb{N}^{*} $, we have :

\begin{aligned} \sum_{k=1}^{n}{\frac{1}{2^{k}k}}&=\sum_{k=1}^{n}{\int_{0}^{\frac{1}{2}}{x^{k-1}\,\mathrm{d}x}}\\&=\int_{0}^{\frac{1}{2}}{\sum_{k=1}^{n}{x^{k-1}}\,\mathrm{d}x}\\ &=\int_{0}^{\frac{1}{2}}{\frac{1-x^{n}}{1-x}\,\mathrm{d}x}\\ \sum_{k=1}^{n}{\frac{1}{2^{k}k}}&=\ln{2}-\int_{0}^{\frac{1}{2}}{\frac{x^{n}}{1-x}\,\mathrm{d}x}\end{aligned}

Since : $$ \int_{0}^{\frac{1}{2}}{\frac{x^{n}}{1-x}\,\mathrm{d}x}\leq\frac{1}{2^{n}}\int_{0}^{\frac{1}{2}}{\frac{\mathrm{d}x}{1-x}}=\frac{\ln{2}}{2^{n}}\underset{n\to +\infty}{\longrightarrow}0 $$

We get that : $$ \sum_{n=1}^{+\infty}{\frac{1}{2^{n}n}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\frac{1}{2^{k}k}}}=\ln{2} $$

Answer 3

$$\frac{1}{1-x}=\sum_{n=1}^{\infty} x^{n-1}$$

$$\int_{0}^{\frac{1}{2}}\frac{1}{(1-x)}dx=\int_{0}^{\frac{1}{2}}\sum_{n=1}^{\infty} x^{n-1}dx$$ $$\log2=\sum_{n=1}^{\infty}\frac{1}{n2^n}$$