Yes that’s correct this is how I started but I’m not sure if it is correctI suspect the issue is that the images are sideways, and not with your handwriting, which is actually very nice. If you're uploading images from a phone, then it's not really your fault...I see a lot of sideways images on forums from mobile users. Is this the question?
a) Find the radius and interval of convergence of:
\[\sum_{n=0}^{\infty}\frac{n(3x-1)^n}{6^n}\]
I did but I’m not sure if it is correctThen use the "limit ratio test": \(\displaystyle \sum a_n\) converges absolutely if \(\displaystyle \lim \left|\frac{a_{n+1}}{a_n}\right|< 1\).
What we want is:
I think I got it now. So IS The radius of convergence 2?Your work is very sloppy. When you replace n with n+1 you must write (n+1). The powers of (3x-1) are both n+1in your work. But this can't be since in one case n stays n and in the other case n becomes (n+1). You were told to take lim |{a_{n+1}/a_{n}|<1, yet you did not take the absolute value nor did you write limit. Also how did you compute that the radius of convergence is 6?
Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?What we want is:
\(\displaystyle \frac{|3x-1|}{6}<1\)
or:
\(\displaystyle \left|x-\frac{1}{3}\right|<2\)
And so the radius of convergence is 2, and the interval of convergence is:
\(\displaystyle \left(\frac{1}{3}-2,\frac{1}{3}+2\right)\)
or:
\(\displaystyle \left(-\frac{5}{3},\frac{7}{3}\right)\)
We have:Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?
RIGHTTT. Makes a lot of sense. Thank you so much for your time.We have:
\(\displaystyle \frac{|3x-1|}{6}<1\)
And we want to get this in the form:
\(\displaystyle |x-a[<r\)
And so we factor out a 3 from the numerator and divide out common factors to get:
\(\displaystyle \frac{\left|x-\dfrac{1}{3}\right|}{2}<1\)
Multiply through by 2:
\(\displaystyle \left|x-\frac{1}{3}\right|<2\)
And so we have it in the form we want, where \(\dfrac{1}{3}\) is the center of the interval of convergence, whose radius is \(2\).
Does that make sense?
Hey. Thanks for helping. But may I know why it diverges for both? . ThanksIf a power series converges for |x-a|<R and diverges for |x-a|>R, then the power series is centred at point a and has radius r=R. In your case a=1/3 and R=2. The interval of convergence is (a-R, a+R) = (-5/3,7/3) . The series diverges for both x=-5/3 and x=7/3 so the interval is open.
When we use the ratio test, we find the limit:Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
When \(\displaystyle x=\frac{7}{3}\) you have L=1 which is the case when the ratio test does not allow to conclude. In this case you replace xHey. Thanks for helping. But may I know why it diverges for both? . Thanks