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Step 2. To develop an estimate for <math> n^{\frac{1}{n}} </math>, Consider the following: | Step 2. To develop an estimate for <math> n^{\frac{1}{n}} </math>, Consider the following: | ||
| + | <math> (x-1) = (x^{1/n}-1)(1+x^{1/n}+x^{2/n}+\ldots + x^{n-1/n}) </math> | ||
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| + | to be continued ... | ||
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Revision as of 12:32, 10 March 2010
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Prof. Alekseenko: Problem #13 is a little bit tricky, so let me give some hints about the solution. The problem specifically directs you to use the Squeeze theorem. The difficult part is to come up with a correct estimate for the sequence.
Part (a) is not very hard. One can use the Bernoulli inequality. I will just indicate the basic steps and you will have to fill in the detail.
Step 1. Show that if $ a^n > b^n $ for some natural number $ n \, $ then $ a > b \, $. You can try to show it by a contradiction.
Step 2. Use a Bernoulli inequality to obtain the following: $ \left( 1+\frac{1}{n} \right)^{n^2} > 1 + n > n $
Conclude that $ 1+\frac{1}{n} > n^{\frac{1}{n^2}} $
Step 3. Use Step 1 again to conclude that $ 1 < n^{\frac{1}{n^2}} $
Step 4. Apply the squeeze theorem.
Part (b) is a bit trickier, however is not very hard. You will have to use some of the results in part (a), not not all of them.
Step 1. Notice that
$ 1 < (n!)^{\frac{1}{n^2}} = (1\cdot 2 \cdot \ldots \cdot n)^{\frac{1}{n^2}} $
$ < (n\cdot n \cdot \ldots \cdot n)^{\frac{1}{n^2}}= n^{\frac{1}{n}} $
Step 2. To develop an estimate for $ n^{\frac{1}{n}} $, Consider the following: $ (x-1) = (x^{1/n}-1)(1+x^{1/n}+x^{2/n}+\ldots + x^{n-1/n}) $
to be continued ...
