Question

Suppose that a function has the following Maclaurin series. Find the radius of convergence.

Answer 1

\[\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{ 3^n n }(x-3)^n\]

Answer 2

I get stuck after changing it into the a(n+1)/an part

Answer 3

Find the radius of convergence of the absolute value. The alternating is guaranteed to converge within this radius.
What do you get for a(n+1)/a(n)

Answer 4

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n+1}}{ 3^nn }(x-3)^n* \frac{ 3^{n+1}(n+1) }{ (-1)^{n+2} (x-3)^{n+1}}\]

Answer 5

and then I cross off the -1^{n+1} and the {x-3}^n

\[\sum_{n=1}^{\infty} \frac{ 1}{ 3^nn }* \frac{ 3^{n+1}(n+1) }{ (-1) (x-3)}\]

Answer 6

\[-3 \sum_{n=1}^{\infty} \frac{ 1 }{ n }*\frac{ n+1 }{ x-3 }\] is this right?

Answer 7

\[\lim_{n\to\infty} \frac{(-1)^{n+2}(x-3)^{n+1}}{ 3^{n+1}(n+1)}* \frac{ 3^n n }{(-1)^{n+1} (x-3)^n}\]

\[\lim_{n\to\infty} \frac{(-1)^{n+2-n-1}(x-3)^{n+1-n}}{ 3^{n+1-n}}* \frac{ n }{(n+1)}\]
\[\lim_{n\to\infty} \frac{(-1)^{1}(x-3)^{1}}{ 3^{1}}* \frac{ n }{(n+1)}\]
\[\left|\frac{(x-3)}{ 3}\right|~\lim_{n\to\infty} \frac{ n }{n+1}\]

Answer 8

\[\left|\frac{(x-3)}{ 3}\right|<1\]
\[|x-3|<3~;~R=3\]

Answer 9

(x-3) = 3, when x=6
(x-3) = -3, when x=0

so the interval of convergence is from 0 to 6, which of you think of as a diameter of a circle ... the radius is half that, which is still 3 :)