Limits

**alikation0** · September 28th 2007, 06:45 AM

Given the sequence $a_n = \frac{7n^2 + 2}{5n^2 - 3},$

1a) Prove $a_n$ converges and then find the lim a.

b) If $\epsilon = \frac{1}{100}$ , give a value of $N$ that will guarantee that $a_n \in V_{\epsilon}(a)$ whenever $n \ge N$ . Then, verify this answer by picking an $n \ge N$ and checking $|a_n - a| < \epsilon$

Hmm, okay so obviously the limit is $\frac{7}{5}$ .. but I know this from basic calculus (since we have the same power in the numerator/denominator). So would I proceed knowing this fact, and then proceeding with the convergence proof with $\epsilon$ .. that is:

Let $\epsilon > 0$ be arbitrary. Then ____ ***have to find is a real number so there exists a natural number $N \in \mathbb{N}$ such that N > _____ ***same thing, have to find by the Archimedean Property. For this $N$ we have $|\frac{7n^2 + 2}{5n^2 - 3} - \frac{7}{5}| \le ... < ... < \epsilon$

OR, would I proceed to showing this converges by using the Monotone Convergence Thm (and thus pretty much making no assumptings as to what it converges to), and showing its bounded and monotone and therefore converges.

For part b, not sure how to do this.

EDIT: Err, wait, it isn't monotone everywhere.

Meh.

**Plato** · September 28th 2007, 07:38 AM

$\left| {\frac{{7n^2 + 2}}{{5n^2 - 3}} - \frac{7}{5}} \right| = \left| {\frac{{31}}{{5\left( {5n^2 - 3} \right)}}} \right| = \frac{{31}}{5}\left| {\frac{1}{{\left( {5n^2 - 3} \right)}}} \right|<br />$
$\frac{{31}}{5}\left| {\frac{1}{{\left( {5n^2 - 3} \right)}}} \right| < \frac{1}{{100}}\quad \Rightarrow \quad \frac{1}{{\left( {5n^2 - 3} \right)}} < \frac{1}{{620}}$

Solve for n.

**alikation0** · October 4th 2007, 01:32 PM

Originally Posted by Plato

$\left| {\frac{{7n^2 + 2}}{{5n^2 - 3}} - \frac{7}{5}} \right| = \left| {\frac{{31}}{{5\left( {5n^2 - 3} \right)}}} \right| = \frac{{31}}{5}\left| {\frac{1}{{\left( {5n^2 - 3} \right)}}} \right|<br />$
$\frac{{31}}{5}\left| {\frac{1}{{\left( {5n^2 - 3} \right)}}} \right| < \frac{1}{{100}}\quad \Rightarrow \quad \frac{1}{{\left( {5n^2 - 3} \right)}} < \frac{1}{{620}}$

Solve for n.

Thanks for the help Plato. I understand what you did, using the definition of convergence. But this doesn't show that it converges. I am assuming that (31/5)*|1/(5n^2 -3)| converges, and then using the |a_n - a| < epsilon to work out part b.

Don't I have to do a series of inequalities, that is:

|(7n^2 +2)/(5n^2 - 3) - (7/5)| = 31/5|1/(5n^2-3)| < stuff here containing N, but not quite sure < whatever < ... = epsilon,

and it is NOW that we can tell it converges? Then I can go ahead and set epsilon to equal 1/100 and solve for n. But I'm not sure how to do that above step.

Thanks for your help.

**Plato** · October 4th 2007, 01:46 PM

$\frac{{31}}{5}\left| {\frac{1}{{5n^2 - 3}}} \right| < \varepsilon \quad \Rightarrow \quad \frac{{ - 5\varepsilon }}{{31}} < \frac{1}{{5n^2 - 3}} < \frac{{5\varepsilon }}{{31}}$

**alikation0** · October 17th 2007, 05:35 PM

Originally Posted by Plato

$\frac{{31}}{5}\left| {\frac{1}{{5n^2 - 3}}} \right| < \varepsilon \quad \Rightarrow \quad \frac{{ - 5\varepsilon }}{{31}} < \frac{1}{{5n^2 - 3}} < \frac{{5\varepsilon }}{{31}}$

Question: does this prove convergence? I thought I understand your logic, trapping n within two epsilons, but my professor says this doesn't prove convergence.

**Plato** · October 18th 2007, 03:47 AM

Originally Posted by alikation0

Question: does this prove convergence? I thought I understand your logic, trapping n within two epsilons, but my professor says this doesn't prove convergence.

Your professor is correct.
You must proceed to solve, finding an N that works.

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