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:











We'll notice that the Sn0.7S_n\rightarrow0.7 as nn\rightarrow\infty and in fact, it is ln2=0.69314718...\ln 2=0.69314718...

Now, S=112+1314+15+....S_\infty=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+....

And 2S=21+2324+2526+2728+29210+...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}+...





So, 2S=S2S_\infty=S_\infty

We know that, SS_\infty is not equal to 00.

Then 2=12=1!!

Please help me solving the mistake.

Note by Fahim Muhtamim
1 year, 8 months ago

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1 vote

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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,r, you can rearrange the terms of the series so that it sums to r.r. Here is a nice reference.

Patrick Corn - 1 year, 8 months ago

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Your mistake is when you put it as 112+(2313)..1-\frac{1}{2}+(\frac{2}{3}-\frac{1}{3})...

The gaps between 23and13,25and15,29and19\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.


Chris Sapiano - 1 year, 8 months ago

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