anyone?

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- February 15th 2010, 04:31 AM #1

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- February 15th 2010, 05:56 AM #2

- February 15th 2010, 06:19 AM #3

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hi nadav3,

For the 1st one....

If the inequality is valid, then we try to use it to prove the following

Proof

As this is then we only need ask if

Now, since |a|<1

Therefore, the extra terms sum to a negative value

and the overall sum is at least 2.

Therefore

if is

Now prove for the first term

- February 15th 2010, 06:39 AM #4

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- February 15th 2010, 06:43 AM #5

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For the 2nd one..

?

If this is true, then the following must be true

.... (1)

We try to use the first statement to prove the second one..

Since these denominators are larger than those in (1),

then these fractions are even smaller than those in (1)

Hence, the inductive step is proven,

meaning that if the inequality holds for any n=k, it holds for the next n=k+1 and the next n=k+2 and the next n=k+3 etc etc,

therefore it only needs to be validated for an initial term.

- February 15th 2010, 06:52 AM #6

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