Question

Can somebody help me prove this?

Answer 1

Prove that the rearrangement of the series \(\sum_{i=1}^\infty \frac{(-1)^{n+1}}{n}=1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} +\dots \) that takes the sum of two positive terms and then a negative term (i.e. \(1+\frac{1}{3} - \frac{1}{2} +\frac{1}{5} + \frac{1}{7} - \frac{1}{4}+\dots\) ) converges and does to a different limit than the original.

Answer 2

hmm both diverge how do u know they converge ?

Answer 3

i know the it converges by the alternating series test
all that is needed is that the absolute value of the terms goes to zero

Answer 4

the original sum is \(\log(2)\)
trying to figure out a nice closed form for the second sum

Answer 5

(1+1/3+1/5+1/7+....)-(1/2+1/4+1/6+...)
eh =(

Answer 6

that is not the rearrangement they are looking for the sum of

Answer 7

hahaha i know i still feel dull about it xD

Answer 8

\[\sum\limits_{n= 1}^{\infty} \dfrac{(-1)^{n+1}}{n} \]
\[\lim\limits_{n\to \infty}\dfrac{1}{n} = 0\\~\\\frac{1}{n}\gt\frac{1}{n+1}\]

By alternating series test the series must converge. Say it converges to \(X\)
\[\sum\limits_{n= 1}^{\infty} \dfrac{(-1)^{n+1}}{n} =X\tag{1}\]

Answer 9

Next consider the rearrangement \[\begin{align}&1+\frac{1}{3} - \frac{1}{2} +\frac{1}{5} + \frac{1}{7} - \frac{1}{4}+\dots = \sum\limits_{n=1}^{\infty} \left(\dfrac{1}{4n-3} + \dfrac{1}{4n-1} -\frac{1}{2n}\right)\\~\\
&= \sum\limits_{n=1}^{\infty} \left(\dfrac{1}{4n-3} \color{blue}{-\dfrac{1}{4n-2}} + \dfrac{1}{4n-1} -\frac{1}{4n} \color{blue}{+ \frac{1}{4n-2}}- \frac{1}{4n} \right)\\~\\
&= \sum\limits_{n=1}^{\infty} \left(\dfrac{1}{4n-3} \color{blue}{-\dfrac{1}{4n-2}} + \dfrac{1}{4n-1} -\frac{1}{4n}\right) +\frac{1}{2}\sum\limits_{n=1}^{\infty} \left( \color{blue}{ \frac{1}{2n-1}}- \frac{1}{2n} \right)\\~\\
&= \sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}+\frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}\\~\\
&= X+\frac{1}{2}X\\~\\
& = \dfrac{3}{2}X
\end{align}\]

Answer 10

:O