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•  » Proving the divergence of the Harmonic series (alternate method(s))

## #1 2011-10-10 05:12:29

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### Proving the divergence of the Harmonic series (alternate method(s))

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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## #2 2011-10-11 01:05:28

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### Re: Proving the divergence of the Harmonic series (alternate method(s))

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 (2011-10-11 01:19:48)

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## #3 2011-10-16 13:20:21

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### Re: Proving the divergence of the Harmonic series (alternate method(s))

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 (2011-10-16 14:56:59)

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## #4 2011-10-21 09:59:13

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### Re: Proving the divergence of the Harmonic series (alternate method(s))

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 (2011-10-21 10:00:10)

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