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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> ( | + | Again, we use the Bernoulli inequality |
| + | <math> \left( 1+\frac{1}{\sqrt{n}} \right)^{n} > 1 + n*\frac{1}{\sqrt{n}} > \sqrt{n} </math> | ||
| + | Which, in turn implies | ||
| + | <math> 1+\frac{1}{\sqrt{n}} > n^{1/2n} </math> | ||
| + | |||
| + | By squaring both sides, | ||
| + | <math> 1+2\frac{1}{\sqrt{n}}+\frac{1}{n} > n^{1/n} </math> | ||
| + | |||
| + | Step 3. Use | ||
| + | <math> 1 < (n!)^{\frac{1}{n^2}} < n^{1/n} < 1+2\frac{1}{\sqrt{n}}+\frac{1}{n} </math> | ||
| + | To conclude that all these goes to 1. | ||
| − | |||
---- | ---- | ||
Revision as of 10:34, 11 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: Again, we use the Bernoulli inequality $ \left( 1+\frac{1}{\sqrt{n}} \right)^{n} > 1 + n*\frac{1}{\sqrt{n}} > \sqrt{n} $
Which, in turn implies $ 1+\frac{1}{\sqrt{n}} > n^{1/2n} $
By squaring both sides, $ 1+2\frac{1}{\sqrt{n}}+\frac{1}{n} > n^{1/n} $
Step 3. Use $ 1 < (n!)^{\frac{1}{n^2}} < n^{1/n} < 1+2\frac{1}{\sqrt{n}}+\frac{1}{n} $ To conclude that all these goes to 1.
