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Prove that:
7/12 < 1/41 + 1/42 + 1/43 + ... + 1/79 + 1/80 < 5/6
Can anyone help ?
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Here's a hint: 1/41 + 1/42 + ... + 1/50 < 1/40 + 1/40 + ... + 1/40 = 10(1/40) = 1/4.
Do a similar thing with three other groups of ten fractions, then look at the four simpler fractions you end up with.
Similarly, you can look at the four groups of ten fractions and for each of them, find a simpler fraction that is less than their sum.
Post again if you need more help.
Why did the vector cross the road?
It wanted to be normal.
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For the purpose of this question, you only need
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Thank you guys, I think that's more than enough,
Last edited by meo_beo (2008-06-03 02:39:15)
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