Quote:
Originally Posted by pdurrant
You're right that it diverges, but your proof isn't very elegant.
Here's the elegant proof.
Divide the series into groups as follows:
1 + (1/2 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/14 + 1/14 + 1/15 + 1/16) + ....
Replace each term in each group with the smallest term in the group. Note that this reduces the sum of the series.
1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + .....
Note that each and every group sums to 1/2
1 + (1/2) + (1/2) + (1/2) + .....
Which clearly diverges to infinity. Our original sequence has a larger sum than this, and so also diverges to infinity.
Q.E.D.
Hamlet53 gets to set the next question.
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I take exception to your not elegant characterization.
Perhaps if I could have put in the standard symbols with a graph to illustrate? But mine is just the standard integral test for infinite series.
Quote:
Originally Posted by Bilbo1967
Sorry, why do you replace with the smallest term in each group? From my decidedly uninformed standpoint it just smacks of cheating.
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Actually I think there is a mistake in Pdurrant's proof.
1+1/2+[1/3 +1/4] +[1/5+1/6+1/7+1/8] + . . . > 1 + 1/2 + (1/4 +1/4) + (1/8+1/8+1/8+1/8) + . . . = 1 +1/2 +1/2+1/2+ . .. diverges.
The point is that each group in the [] in the original series is larger than the group in the () in the second series; 1/3+1/4 > 1/4 +1/4.