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Hi, this one may seem trivial but I really need help. Prove that:
1/11 + 1/12 + 1/13 + .... + 1/80 < 1.5
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Well, I think you get the inequality sign for the question wrong.
Assuming what you type is
Observe that
Therefore,
Last edited by coffeeking (2009-02-17 16:02:16)
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How can you say that 1/11 + 1/12+ ....+ 1/40 > 4(1/40) ?
It would be ok to say : 1 + 1/2 + 1/3 + 1/4 + ....+ 1/11 + ....+ 1/40 > 40(1/40). But here the sequence starts at 1/11 so I'm afraid there's something wrong with your reasoning.
Last edited by meo_beo (2009-02-17 17:46:28)
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ok, here's my version of proof
sorry if I made any mistakes
Last edited by Dragonshade (2009-02-17 18:35:02)
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Oh my gosh... Think I need more coffee you are right meo_beo, I have made a mistake sorry for the confusion.
Anyway, Dragonshade has a great proof. However if I observed correctly this time round, I think the last line of your proof should be
Since from
there are a total of terms.Offline
There are 68 terms because the term corresponding to 1/45 is 1/45. However, you can't include 1/45 in the sequence twice.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Coffeeking's method works with a bit of a tweak.
Group the first ten terms together, then the next twenty and then the other forty.
This gives that the sum is greater than 10(1/20) + 20(1/40) + 40(1/80) = 1/2+1/2+1/2 = 1.5.
Why did the vector cross the road?
It wanted to be normal.
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Dragonshades method looks to be the best so far although
is logically the wrong way round. Youre supposed to start with
and work backwards to
.I was thinking maybe you could say that
This is easily justified graphically.
Hence
which sets an even greater lower bound for the sum.
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Right then, folks. To prove
you only need to prove that
for all positive real numbers (the sum follows by adding terms piecewise).First, we prove this:
Proof:
In particular,
for allOffline
Pretty good, Jane!
I also like how you can use similar reasoning to say that the sum is less than ln(80/10) ≈ 2.07944.
So we know that the answer is somewhere in ( ln(81/11), ln(8) ).
Now I'm wondering if there's some way of showing that the answer is close to the midpoint of the interval, and improving the bounds that way.
Why did the vector cross the road?
It wanted to be normal.
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Well, you can improve on the accuracy by using simple tricks. For example, leaving out the first term,
If you want even better,
etc.
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