I can't find any mistake here

Let, $S_\infty=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+....$

By calculating, we find that:

$S_1=1$

$S_2=\frac{1}{2}$

$S_3=0.833333...$

$S_4=0.5833333...$

$S_5=0.78333...$

$S_6=0.616666...$

$S_7=0.759523809...$

$S_8=0.634523809...$

$S_9=0.7445492...$

$S_{10}=0.64563492...$

We'll notice that the $S_n\rightarrow0.7$ as $n\rightarrow\infty$ and in fact, it is $\ln 2=0.69314718...$

Now, $S_\infty=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+....$

And $2S_\infty=2-1+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-\frac{2}{6}+\frac{2}{7}-\frac{2}{8}+\frac{2}{9}-\frac{2}{10}+...$

$=1+\frac{2}{3}-\frac{1}{2}+\frac{2}{5}-\frac{1}{3}+\frac{2}{7}-\frac{1}{4}+\frac{2}{9}-\frac{1}{5}+...$

$=1-\frac{1}{2}+(\frac{2}{3}-\frac{1}{3})-\frac{1}{4}+(\frac{2}{5}-\frac{1}{5})-...$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+...$

$=S_\infty$

So, $2S_\infty=S_\infty$

We know that, $S_\infty$ is not equal to $0$ .

Then $2=1$ !!

Please help me solving the mistake.

#Algebra

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Comments

Rearranging the terms of a sum that is not absolutely convergent may change the value of the sum. In fact, there is a wonderful result that if a series is convergent but not absolutely convergent, then for any real number $r,$ you can rearrange the terms of the series so that it sums to $r.$ Here is a nice reference.

Patrick Corn - 1 year, 6 months ago

Your mistake is when you put it as $1-\frac{1}{2}+(\frac{2}{3}-\frac{1}{3})..$ .

The gaps between $\frac{2}{3} and \frac{1}{3}, \frac{2}{5}and \frac{1}{5},\frac{2}{9} and \frac{1}{9}$ . Increase each time. With this logic you can prove that there are twice as many even numbers as odd numbers.

$1,3,2,5,7,4,9,11,6..$

Chris Sapiano - 1 year, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$