Jbreezy said:
Homework Statement
Say that you were using ration test for
##
\sum_{n=1}^\infty\frac{(-1)^{n+1} (x-4)^n}{n9^n!}\
##
Homework Equations
The Attempt at a Solution
You take the limit of the above you will get
##\frac {1}{9} |x-4|##Book says radius of convergence is 9? Is this done by using the ratio test and making an inequality?
For others who might be wondering, yes, this is correct. The "ratio test" says that a series , [itex]\sum a_n[/itex] converges (absolutely) as long as
[tex]\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|< 1[/tex]
Here, that ratio is [itex][|x- 4|^{n+1}]/[(n+1)9^{n+1}][n 9^n]/[|x- 4|^n= [n/(n+1)]|x- 4|/9[/itex]. (I have ignored the "[itex](-1)^{n+1}[/itex]" because of the absolute value.)
The limit as n goes to infinity is, as Jbreezy said, |x- 4|/9. From |x- 4|/9< 1 we obviously get |x- 4|< 9 and so -9< x- 4< 9. The radius of convergence, the distance from the center point, x= 4. to each end of the interval of convergence is 9. So, adding 4 to each part,-9+ 4= -5< x< 13= 9+ 4.
Your "error" is trying to incorporate the absolute value where it does not belong.
Now, the series might or might (or might converge but not "absolutely") at x= -5 and x= 13. Those you would have to check separately.
##\frac {1}{9} |x-4| <1##
##|x-4| < 9##
No?
OK then the book says in the next line the
##-5< x< 13##
How did they get this? I thought that they maybe said if R = 9 then put ,-9 and 9 in the inequality. So,
##|(-9)-4| = -13## and ## |(9)-4| = 5##
But that doesn't work because it is backwards theirs was -5 and 13 not 13 and -5
What is going on. If you can elaborate the idea of radius of convergence and how they determined the interval please!