Solving Ratio Test Limit with x and -3

In summary: Sorry for all the mistakes, but I don't see the problem. If you use the ratio test, you get lim|a(n+1)/a(n)| = |x/3| for x not 0. So if that is less than 1, you have convergence, and if that is more than 1, you have divergence.
  • #1
Dannbr
9
0
[tex]\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}[/tex]
[tex]\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}[/tex]

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

[tex] = \left| {\frac{x}{3}} \right|[/tex]

could someone help me out with what I am not seeing

Thanks

 
Physics news on Phys.org
  • #2
Removed some extraneous stuff.
Dannbr said:
[tex]\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}[/tex]

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

[tex] = \left| {\frac{x}{3}} \right|[/tex]

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.
 
  • #3
here it is:

[tex]\sum\limits_{n = 0}^\infty {\left[ {\frac{1}{{{{\left( { - 3} \right)}^n}}} - 1} \right]} {x^n}[/tex]
 
  • #4

Homework Statement



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

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


Homework Equations





The Attempt at a Solution



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

Using PFD I came up with...

[tex]\frac{3}{{x + 3}} + \frac{1}{{x - 1}}[/tex]


then set it to the form...

=[tex]\frac{1}{{1 - \left( { - \frac{x}{3}} \right)}} - \frac{1}{{1 - x}}[/tex]

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

[tex]\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}[/tex]

the book says Interval of convergence:

[tex]\left| {\frac{x}{3}} \right| < 1and\left| x \right| < 1 \Rightarrow ( - 1,1)[/tex]

I understand now it diverges at (-1,1) because it is (-3,3) is larger
 
  • #5
Dannbr said:

Homework Statement



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

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


Homework Equations





The Attempt at a Solution



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

Using PFD I came up with...

[tex]\frac{3}{{x + 3}} + \frac{1}{{x - 1}}[/tex]
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.
Dannbr said:
then set it to the form...

=[tex]\frac{1}{{1 - \left( { - \frac{x}{3}} \right)}} - \frac{1}{{1 - x}}[/tex]

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

[tex]\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}[/tex]

the book says Interval of convergence:

[tex]\left| {\frac{x}{3}} \right| < 1and\left| x \right| < 1 \Rightarrow ( - 1,1)[/tex]

I understand now it diverges at (-1,1) because it is (-3,3) is larger
 
  • #6
yea, you are right. I posted the wrong problem g(x)

g(x) should be ...


[tex]g(x) = \frac{{4x}}{{{x^2} + 2x - 3}}[/tex]


Sorry.. this whole thing is screwed up
 
  • #7
I put in the correct g(x).
Dannbr said:

Homework Statement



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

[tex]g(x) = \frac{4x}{x^2 + 2x - 3}[/tex]


Homework Equations





The Attempt at a Solution



[tex]g(x) = \frac{4x}{x^2 + 2x - 3}[/tex]


Using PFD I came up with...

[tex]g(x) = \frac{3}{{x + 3}} + \frac{1}{{x - 1}}[/tex]
So we should be good at least to here, and the next one or maybe two look correct.
Dannbr said:
then set it to the form...

=[tex]\frac{1}{{1 - \left( { - \frac{x}{3}} \right)}} - \frac{1}{{1 - x}}[/tex]

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

[tex]\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}[/tex]

the book says Interval of convergence:

[tex]\left| {\frac{x}{3}} \right| < 1and\left| x \right| < 1 \Rightarrow ( - 1,1)[/tex]

I understand now it diverges at (-1,1) because it is (-3,3) is larger
 
  • #8
In this line,
[tex]\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}[/tex]

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

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.
 

Related to Solving Ratio Test Limit with x and -3

1. What is the purpose of solving a ratio test limit with x and -3?

The ratio test is a method used to determine the convergence or divergence of a series. By solving the limit with x and -3, we can determine if the series converges or diverges at that specific point.

2. How do you solve a ratio test limit with x and -3?

To solve a ratio test limit with x and -3, we first need to take the absolute value of the series and then apply the limit as x approaches -3. If the resulting limit is less than 1, the series converges. If it is greater than 1, the series diverges. If it is equal to 1, the test is inconclusive and another method must be used.

3. What is the significance of -3 in solving a ratio test limit?

The number -3 represents the value at which we are evaluating the series. By plugging in this specific value for x, we can determine the behavior of the series at that point.

4. Can a ratio test limit with x and -3 be used for any series?

No, the ratio test is only applicable for certain types of series, such as geometric series and series with factorial terms. It is important to check the conditions for the ratio test before attempting to use it.

5. What is the advantage of using the ratio test over other tests?

The ratio test is a powerful tool because it can determine the convergence or divergence of a series even if it is not an alternating series or a series with strictly decreasing terms. This makes it a versatile test that can be used for a wide range of series.

Similar threads

Interval of convergence of power series from ratio test
    • Calculus and Beyond Homework Help
Replies
2
Views
330
How to show that these sequences are summable?
    • Calculus and Beyond Homework Help
Replies
1
Views
420
Fourier series of this scale and shift of triangle wave
    • Calculus and Beyond Homework Help
Replies
1
Views
305
On the ratio test for power series
    • Calculus and Beyond Homework Help
Replies
2
Views
774
Prove that this series converges uniformly on R
    • Calculus and Beyond Homework Help
Replies
4
Views
495
Evaluate the limit of this harmonic series as n tends to infinity
    • Calculus and Beyond Homework Help
Replies
17
Views
730
Show that the given function is decreasing
    • Calculus and Beyond Homework Help
Replies
9
Views
775
Show that the limit (1+z/n)^n=e^z holds
    • Calculus and Beyond Homework Help
Replies
6
Views
531
Solution verification of ODE using Frobenius' method
    • Calculus and Beyond Homework Help
Replies
3
Views
556
Find limit involving square of sine
    • Calculus and Beyond Homework Help
Replies
9
Views
1K

Hot Threads

Recent Insights

Back
Top