Convergence of Series: Finding x for Convergence | Homework Statement

[DiamondV]

Homework Statement

For which number x does the following series converge:
http://puu.sh/lp50I/3de017ea9f.png

Homework Equations

abs(r) is less than 1 then it is convergent. r is what's inside the brackets to the power of n

The Attempt at a Solution

I did the question by using the stuff in relevant equations above eventually leading to a quadratic which I solved and got two answers x is 1 and x is 3. However the solutions given to us by our instructor says
http://puu.sh/lp5be/208cccba8f.png
Why is my answer not the same

 

[LCKurtz]

Homework Statement

For which number x does the following series converge:
http://puu.sh/lp50I/3de017ea9f.png

Homework Equations

abs(r) is less than 1 then it is convergent. r is what's inside the brackets to the power of n

The Attempt at a Solution

I did the question by using the stuff in relevant equations above eventually leading to a quadratic which I solved and got two answers x is 1 and x is 3. However the solutions given to us by our instructor says
http://puu.sh/lp5be/208cccba8f.png
Why is my answer not the same

You haven't shown your work, so how can we tell? How did you solve$$
\left |\frac x {3-2x}\right | < 1\text{ ?}$$

 

[Buzz Bloom]

Hi Diamond:

You made a mistake interpreting the absolute value.

abs(a/b) < 1​

means

abs(a) < abs(b) and
abs(a) < -abs(b) and
-abs(a) < abs(b) and
-abs(a) < -abs(b).​

Regards,
Buzz

 

[DiamondV]

You haven't shown your work, so how can we tell? How did you solve$$
\left |\frac x {3-2x}\right | < 1\text{ ?}$$

Hi Diamond:

You made a mistake interpreting the absolute value.

abs(a/b) < 1​

means

abs(a) < abs(b) and
abs(a) < -abs(b) and
-abs(a) < abs(b) and
-abs(a) < -abs(b).​

Regards,
Buzz

Sorry about that. Here's my working:
http://puu.sh/lpa5y/d2810ed6ed.jpg

 

[LCKurtz]

Multiply that 4th line up by -1 and divide by 3 to get $x^2-4x+3>0$, or $(x-3)(x-1)>0$. Look at the signs of the factors to see where the product is positive. Simplifying things lessens chances for errors.

 

[DiamondV]

Multiply that 4th line up by -1 and divide by 3 to get $x^2-4x+3>0$, or $(x-3)(x-1)>0$. Look at the signs of the factors to see where the product is positive. Simplifying things lessens chances for errors.

Don't see how that explains the answer given to me in the solutions including the infinity part?

 

[Buzz Bloom]

Hi Diamond:

I think the squaring of the inequality and solving the quadratic makes the problem more complicated than necessary, and easier to make a mistake. I suggest solving each of the 4 simple inequalities I mentioned in my post. Actually, two of the four inequalities are redundant, so you only need to solve to get two different result sets of values of x. The union of the the two sets of x values is the answer you want.

Regards,
Buzz

 

[LCKurtz]

Multiply that 4th line up by -1 and divide by 3 to get $x^2-4x+3>0$, or $(x-3)(x-1)>0$. Look at the signs of the factors to see where the product is positive. Simplifying things lessens chances for errors.

Don't see how that explains the answer given to me in the solutions including the infinity part?

The $x$ values that work in that inequality are the $x$ values that solve your original problem. Each factor is either positive or negative, depending on the value of $x$. One factor changes sign when $x=3$ and the other at $x = 1$ (which are the values you found). Draw a number line and mark 1 and 3 on it. Those two points are where the product is $0$. Everywhere else the factors are either + or -. Look at the signs of the factors on the three intervals. If you find $-\cdot -$ or $+\cdot +$ you know the product is positive on that interval.

 

[SammyS]

The $x$ values that work in that inequality are the $x$ values that solve your original problem. Each factor is either positive or negative, depending on the value of $x$. One factor changes sign when $x=3$ and the other at $x = 1$ (which are the values you found). Draw a number line and mark 1 and 3 on it. Those two points are where the product is $0$. Everywhere else the factors are either + or -. Look at the signs of the factors on the three intervals. If you find $-\cdot -$ or $+\cdot +$ you know the product is positive on that interval.

Yes! This is one of the classic ways to solve such a problem !