You are not logged in.
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)
Signature line:
I wish a had a more interesting signature line.
Offline
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)
Signature line:
I wish a had a more interesting signature line.
Offline
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)
Signature line:
I wish a had a more interesting signature line.
Offline
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.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Last edited by reconsideryouranswer (2011-10-20 11:00:10)
Signature line:
I wish a had a more interesting signature line.
Offline