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This is humble proof about harmonic series on my own.

1 + 1/2 + 1/3 + 1/4 + 1/5.......

= 1 + (1 - 1/2) + {(1 - 1/2) - (1/2 - 1/3)} + {(1 - 1/2) - (1/2 - 1/3) - (1/3 - 1/4)} +...

= 1 + (1/2)n - (1/6)(n-1) - (1/12)(n-2) - (1/20)(n-3)+.....

= 1 + (1/2 - 1/6 - 1/12 - 1/20 -.....)n + (1/6 + 1/6 + 3/20 + 4/30 +....)

and n=>unlimited, therefore harmonic series is divergent.

(add explanation)

1 + 1/2 + 1/3 + 1/4 + 1/5.......

= 1

 + (1 - 1/2)

 + (1 - 1/2) - (1/2 - 1/3) 

 + (1 - 1/2) - (1/2 - 1/3) - (1/3 - 1/4)

 + (1 - 1/2) - (1/2 - 1/3) - (1/3 - 1/4) - (1/4 - 1/5)..
 .
 .

= 1 + (1/2)n - (1/6)(n-1) - (1/12)(n-2) - (1/20)(n-3)+.....

= 1 + (1/2 - 1/6 - 1/12 - 1/20 -.....)n + (1/6 + 1/6 + 3/20 + 4/30+....)

and n=>unlimited, harmonic series is divergent, too.

but I think it is out of my capability to identify the validity of this expansion

so, give me a correct judgement about my own speculation.

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    $\begingroup$ Please write clearly . you can do it by latex ...or online latex editor ....your question is my favorite question ,but I am not able to see what you write ! $\endgroup$ – Khosrotash Aug 1 '17 at 15:12
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    $\begingroup$ Frankly speaking, you are making the simple harmonic series something very complicated. Calculus obeys simple rules that everybody understands, but your speculations don't seem to obey these rules. For example, you are computing with quantities that have never been defined such as $1 + 1/2 + 1/3 + \cdots $. Use only well defined expressions. $\endgroup$ – Gribouillis Aug 1 '17 at 15:16
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Your proof, as I understand it, is not valid, as it relies on rearranging the series elements.

In fact, for a large class of series you can find a rearrangement such that it converges to an arbitrary element. This is the content of the Riemann series theorem.

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  • $\begingroup$ IMHO you've commented in what should be an answer field. What approach is valid? Can you show an example of a divergent series, upon rearrangement becomes convergent? As it stands, this belongs as a comment below the question. $\endgroup$ – Namaste Aug 1 '17 at 15:44
  • $\begingroup$ Um, the original series only involves positive terms.... if it converged, it would have to converge absolutely... $\endgroup$ – Simply Beautiful Art Aug 1 '17 at 16:34
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    $\begingroup$ Yes, but the proprosed proof contains negative terms. The sequence in the proof potentially does not converge absolutely. This is not a statement about the original series. Rather it illustrates a problem with the proposed proof. $\endgroup$ – Reiner Martin Aug 1 '17 at 16:35
  • $\begingroup$ So, do you say my proposed expansion can be reformed to the convergent series ? Could you show me the rearrangement that is convergent, using the proposed expansion ? As you say, though it contains opposite terms, I think it wouldn't be the absolute answer to be able to call the proposed proof conditionally convergent. In fact, thank for your sincere interest and answer. But, sorry. It's not easy to agree with your opinion... $\endgroup$ – Clerk Park Aug 2 '17 at 15:08
  • $\begingroup$ No, sorry, I was saying less than that. All I was trying to say is that you have to be careful when you make a statement about series (which are not absolutely convergent) where you are rearranging terms. I might have misunderstood your proof, though. $\endgroup$ – Reiner Martin Aug 2 '17 at 22:02

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