Solving Ratio Test Limit with x and -3

[Dannbr]

\begin{array}{l}<br /> \\<br /> \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|<br /> \end{array}
\begin{array}{l}<br /> \\<br /> <br /> \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|<br /> \end{array}

This is as far as i can get I have an answer key which says it simplifies down to..

= \left| {\frac{x}{3}} \right|

could someone help me out with what I am not seeing

Thanks

 

[Mark44]

Removed some extraneous stuff.

\begin{array}{l}<br /> \\<br /> \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|<br /> \end{array}

This is as far as i can get I have an answer key which says it simplifies down to..

= \left| {\frac{x}{3}} \right|

could someone help me out with what I am not seeing

Thanks

You're apparently working with a series. Can you provide the series itself? I want to make sure that you're working with the right expressions in the ratio test.

 

[Dannbr]

here it is:

\sum\limits_{n = 0}^\infty {\left[ {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right]} {x^n}

 

[Dannbr]

Homework Statement

Find a power series for the function, centered at c, and determine the interval of convergence.

g(x) = \frac{{3x - 8}}{{3{x^2} + 5x - 2}}

Homework Equations

The Attempt at a Solution

g(x) = \frac{{3x - 8}}{{3{x^2} + 5x - 2}}

Using PFD I came up with...

\frac{3}{{x + 3}} + \frac{1}{{x - 1}}


then set it to the form...

=\frac{1}{{1 - \left( { - \frac{x}{3}} \right)}} - \frac{1}{{1 - x}}

\begin{array}{l}<br /> \\<br /> = \sum\limits_{n = 0}^\infty {\left[ {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right]} {x^n}<br /> \end{array}

\begin{array}{l}<br /> \\<br /> \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|<br /> \end{array}

the book says Interval of convergence:

\left| {\frac{x}{3}} \right| &lt; 1and\left| x \right| &lt; 1 \Rightarrow ( - 1,1)

I understand now it diverges at (-1,1) because it is (-3,3) is larger

 

[Mark44]

Homework Statement

Find a power series for the function, centered at c, and determine the interval of convergence.

g(x) = \frac{{3x - 8}}{{3{x^2} + 5x - 2}}

Homework Equations

The Attempt at a Solution

g(x) = \frac{{3x - 8}}{{3{x^2} + 5x - 2}}

Using PFD I came up with...

\frac{3}{{x + 3}} + \frac{1}{{x - 1}}

Your decomposition is incorrect. 3x² + 5x - 2 = (3x - 1)(x + 2). So the denominators in your partial fractions won't be x + 3 and x - 1.

then set it to the form...

=\frac{1}{{1 - \left( { - \frac{x}{3}} \right)}} - \frac{1}{{1 - x}}

\begin{array}{l}<br /> \\<br /> = \sum\limits_{n = 0}^\infty {\left[ {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right]} {x^n}<br /> \end{array}

\begin{array}{l}<br /> \\<br /> \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|<br /> \end{array}

the book says Interval of convergence:

\left| {\frac{x}{3}} \right| &lt; 1and\left| x \right| &lt; 1 \Rightarrow ( - 1,1)

I understand now it diverges at (-1,1) because it is (-3,3) is larger

 

[Dannbr]

yea, you are right. I posted the wrong problem g(x)

g(x) should be ...


g(x) = \frac{{4x}}{{{x^2} + 2x - 3}}


Sorry.. this whole thing is screwed up

 

[Mark44]

I put in the correct g(x).

Homework Statement

Find a power series for the function, centered at c, and determine the interval of convergence.

g(x) = \frac{4x}{x^2 + 2x - 3}

Homework Equations

The Attempt at a Solution

g(x) = \frac{4x}{x^2 + 2x - 3}


Using PFD I came up with...

g(x) = \frac{3}{{x + 3}} + \frac{1}{{x - 1}}

So we should be good at least to here, and the next one or maybe two look correct.

then set it to the form...

=\frac{1}{{1 - \left( { - \frac{x}{3}} \right)}} - \frac{1}{{1 - x}}

\begin{array}{l}<br /> \\<br /> = \sum\limits_{n = 0}^\infty {\left[ {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right]} {x^n}<br /> \end{array}

\begin{array}{l}<br /> \\<br /> \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|<br /> \end{array}

the book says Interval of convergence:

\left| {\frac{x}{3}} \right| &lt; 1and\left| x \right| &lt; 1 \Rightarrow ( - 1,1)

I understand now it diverges at (-1,1) because it is (-3,3) is larger

 

[Mark44]

In this line,
\begin{array}{l}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}\left( {\frac{1}{{{{\left( { - 3} \right)}^{n + 1}}}} - 1} \right)}}{{{x^n}\left( {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\left( {\frac{{{{\left( x \right)}^n}\left( x \right)}}{{{{\left( { - 3} \right)}^n}\left( { - 3} \right)}} - \frac{{{{\left( x \right)}^n}\left( x \right)}}{1}} \right) \bullet \left( {\frac{{{{\left( { - 3} \right)}^n}}}{{{x^n}}} - \frac{1}{{{x^n}}}} \right)} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{x}{{\left( { - 3} \right)}} - x} \right|\end{array}

in the very first limit I would cancel the x factors. This leaves you with
|x|\lim_{n \to \infty}\left| \frac{\frac{1}{(-3)^{n+1}} - 1}{\frac{1}{(-3)^n} - 1}\right|

Instead of multiplying stuff out as you did, I would combine the two terms in the numerator and the two in the denominator, and see if I could simplify that. Another approach is to multiply by (-3)^n over itself. This limit should come out to 1/3.