2

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.

CHAMSI
  • 9,222

3 Answers3

6

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

CHAMSI
  • 9,222
  • I understand this, however does this assume we already know that the sum of the series is ln2 to begin with? –  May 02 '20 at 15:26
  • No, this is more like we pulled $\ln(2)$ out of thin air and then showed that it equals the sum. There is hidden work in the background that gave the number $\ln(2)$ – QC_QAOA May 02 '20 at 15:28
  • @math2020 see the update. – CHAMSI May 02 '20 at 15:34
  • @QC_QAOA what about now, you can better see where the $ \ln{2} $ came from. – CHAMSI May 02 '20 at 15:37
  • 1
    Its more clear now, but there was nothing wrong logically with the previous proof. This way might be better for showing the concepts now. +1 – QC_QAOA May 02 '20 at 15:39
5

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

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

QC_QAOA
  • 12,277
0

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

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

E.H.E
  • 23,590