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[[ HomeworkDiscussionsMA341Spring2010 | go back to the Discussion Page ]] | [[ HomeworkDiscussionsMA341Spring2010 | go back to the Discussion Page ]] | ||
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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 <math> a^n > b^n </math> for some natural number <math> n \, </math> then <math> a > b \, </math>. You can try to show it by a contradiction. | ||
| + | |||
| + | Step 2. Use a Bernoulli inequality to obtain the following: | ||
| + | <math> \left( 1+\frac{1}{n} \right)^{n^2} > 1 + n > n </math> | ||
| + | |||
| + | Conclude that | ||
| + | <math> 1+\frac{1}{n} > n^{\frac{1}{n^2}} </math> | ||
| + | |||
| + | Step 3. Use Step 1 again to conclude that | ||
| + | <math> 1 < n^{\frac{1}{n^2}} </math> | ||
| + | |||
| + | 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 | ||
| + | |||
| + | <math> 1 < (n!)^{\frac{1}{n^2}} = (1\cdot 2 \cdot \ldots \cdot n)^{\frac{1}{n^2}}</math> | ||
| + | |||
| + | <math> < (n\cdot n \cdot \ldots \cdot n)^{\frac{1}{n^2}}= n^{\frac{1}{n}}</math> | ||
| + | |||
| + | Step 2. To develop an estimate for <math> n^{\frac{1}{n}} </math>, Consider the following: | ||
| + | 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. | ||
| + | |||
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Latest revision as of 10:36, 11 March 2010
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go back to the Discussion Page
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.
