Need help with this series

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hajfajv

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So I translated the assignment to the following:

For every [imath]x \in \Reals[/imath] consider this infinite series
[math](*) \displaystyle\sum_{n=0}^\infin (-2x+3)^n[/math]Show that [imath](*)[/imath] is convergent if and only if [math]x \in ]1,2[.[/math]
That's the assignment. However I don't see anywhere in our book or from classes that we have the tools to deal with [imath](-2x+3)^n[/imath]
What I do have is a lot if we have a series like [math]s_n = ak^n[/math] and it is either infinite or finite.
Also why is this only true if the interval is between 1 and 2? if [imath]x = -1[/imath] we get [imath](2+3)^n[/imath] which surely tends towards infinity.
What I tried is [math](*) \displaystyle\sum_{n=0}^\infin ak^n[/math][imath]|k| < 1 \Leftrightarrow -1 < k < 1[/imath]
[imath]s = \frac{a}{1-k}[/imath]
so in my mind it would be
[imath]s = \frac{2}{1-x}[/imath]
so for the interval [imath]s = \frac{2}{1-1}[/imath] which is not good
and [math]s = \frac{2}{1-2} = -1[/math] which is also not living up to the [imath]|k| < 1 \Leftrightarrow -1 < k < 1[/imath]
 
Last edited:
This does make more sense, also it gives result.
However I still don't understand the "if and only if" statement in the assignment, as it doesn't seem to be true.
[math]\frac{1}{1-(|-2*10+3|)} < 1[/math][math]\frac{1}{1-(|-2*(-10)+3|)} < 1[/math]It doesn't matter what x is.
 
hajfajv said:
However I still don't understand the "if and only if" statement in the assignment, as it doesn't seem to be true.
[math]\frac{1}{1-(|-2*10+3|)} < 1[/math][math]\frac{1}{1-(|-2*(-10)+3|)} < 1[/math]It doesn't matter what x is.
Click to expand...

But convergence is not the same thing as the formula for the sum having a value, or being less than 1! Convergence depends on |k| being less than 1.

Take the geometric series with a=1 and k=3, namely 1+3+9+27+..., which clearly is not convergent. Yet you can calculate [math]s=\frac{a}{1-k}=\frac{1}{1-3}=-\frac{1}{2}[/math]
This number is simply irrelevant.
 
Dr.Peterson said:
But convergence is not the same thing as the formula for the sum having a value, or being less than 1! Convergence depends on |k| being less than 1.

Take the geometric series with a=1 and k=3, namely 1+3+9+27+..., which clearly is not convergent. Yet you can calculate [math]s=\frac{a}{1-k}=\frac{1}{1-3}=-\frac{1}{2}[/math]
This number is simply irrelevant.
Click to expand...
oh, thanks! :)
 
hajfajv said:
This does make more sense, also it gives result.
However I still don't understand the "if and only if" statement in the assignment, as it doesn't seem to be true.
[math]\frac{1}{1-(|-2*10+3|)} < 1[/math][math]\frac{1}{1-(|-2*(-10)+3|)} < 1[/math]It doesn't matter what x is.
Click to expand...
First you have to determine whether the sum converge. If it does then it will converge to [imath]\frac{a}{1-k}[/imath]
 
hajfajv said:
So I translated the assignment to the following:
For every [imath]x \in \Reals[/imath] consider this infinite series
[math](*) \displaystyle\sum_{n=0}^\infin (-2x+3)^n[/math]Show that [imath](*)[/imath] is convergent if and only if [math]x \in ]1,2[.[/math]
Click to expand...
An infinite geometric series, [imath]\sum\limits_{n = 1}^\infty {{r^n}} [/imath] converges iff [imath]|r|<1[/imath].
Your notation [imath]\left]1,2\right[[/imath] is equivalent to the open interval [imath](1,2)=\{x:1<x<2\}[/imath].
Also note that [imath]|-2x+3|=|2x-3|[/imath] thus [imath]|2x-3|<1 \Leftrightarrow x\in \left]1,2\right[ [/imath]

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]
 
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