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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

Without using either of the two specific methods shown at this

Wikipedia site (in the top first few pages), use an alternate

method to prove the divergence of the Harmonic series:

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

I want everyone to see this, so I did not hide it:

Prove that the Harmonic series is divergent.

Let S = the Harmonic series.

Assume S has a finite value.

S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...

S = (1 + 1/2) + (1/3 + 1/4) + (1/5 + 1/6) + ...

> (2/2) + (2/4) + (2/6) + ...

= 1 + 1/2 + 1/3 + ...

= S

But this states that S > S, which is impossible,

because S is assumed to be finite.

So, this is a contradiction, and the Harmonic series

is divergent.

*Last edited by reconsideryouranswer (2011-10-10 02:19:48)*

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

I will start with:

Again, assume S is a finite sum.

S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + ...

S = 1 + (1/2 + 1/3) + (1/4 + 1/5) + (1/6 + 1/7) + (1/8 + 1/9) + ...

S = 1 + (5/6) + (9/20) + (13/42) + (17/72) + ...

*Last edited by reconsideryouranswer (2011-10-15 15:56:59)*

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

Another way:

Assume S is a finite sum.

Group it as:

S = 1 + [(1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7) + (1/8 + 1/9 + 1/10) + ...] > **

1 + [3(1/3) + 3(1/6) + 3(1/9) + ...] =

1 + [1 + 1/2 + 1/3 + ...] =

1 + S

But this has S > 1 + S.

This is impossible if S is to have a finite value.

So this is a contradiction, and therefore S must

not have a finite sum.

That is, S is divergent.

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*Last edited by reconsideryouranswer (2011-10-20 11:00:10)*

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