Question

Find the interval of convergence of:

Answer 1

\[\sum_{n=0}^{\infty}n4^{-n}(x-4)^{n}\]

Answer 2

\[
\frac{a_{n+1}}{a_n}=\left | \frac{(n+1) 4^{-n-1}
(x-a)^{n+1}}{n 4^{-n}
(x-a)^n}\right | =\frac{(n+1)
\left| x-a\right| }{4 n}

\]

Answer 3

\[
\lim_{n\to \infty}

\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\left | \frac{(n+1) 4^{-n-1}
(x-4)^{n+1}}{n 4^{-n}
(x-4)^n}\right | =\lim_{n\to \infty}\frac{(n+1)
\left| x-4\right| }{4 n}=\frac {|x-4|}{4}<1\\
|x-4| < 4 \\
-4< x -4< +4\\
0< x <8


\]
We have to examine the end points separately

Answer 4

It is easy to see that the series does not converge for x=0 or x =4

Answer 5

I'm still a bit confused, what is the interval of convergence then?

Answer 6

I am confused where you had "does not converge for x=4", as arent we examining the end points 0 and 8?

Answer 7

From my checking of endpoints, the series don't converge at the point x=0 and x=8. So the interval of convergence is 0<x<8 ?

Answer 8

@eliassaab