Question

Find the radius and interval of convergence of the series:

Answer 1

\[\sum_{n=1}^{\infty}(n(x-4)^n)/(n^3+1)\]

Answer 2

Ok, so to find interval and radius of convergence, one of the best ways is to use the ratio test:\[\lim_{n \rightarrow \infty}\left| a _{n+1}/a_{n} \right|<1\] and solve for what values of x make this true

Answer 3

so I substitute for n= n+1, so how do I solve for x with n's every where?

Answer 4

the "n"s will disappear with the limit. i will post the complete solution explaining all steps in a second after i type it

Answer 5

for this problem\[a _{n}=n(x-4)^n/(n^3+1)\]
so our limit becomes\[\lim_{n \rightarrow \infty}\left| [(n+1)(x-4)^{n+1}/((n+1)^3+1)]/[(n)(x-4)^{n}/(n^3+1) \right|<1\]
canceling out our terms, that simplifies to:
\[\lim_{n \rightarrow \infty}\left| [(n+1)(x-4)/(n^3+3n^2+3n+2)/[n/(n^3+1)] \right|<1\]
we can pull out the (x-4) from the limit as it has no bearing on evaluating the limit:
\[|x-4|\lim_{n \rightarrow \infty}\left| [(n+1)/(n^3+3n^2+3n+2)/[n/(n^3+1)] \right|<1\]
in order to save time typing, the limit evaluates as such:
\[\lim_{n \rightarrow \infty}\left| [(n+1)/(n^3+3n^2+3n+2)/[n/(n^3+1)] \right|=\]\[\lim_{n \rightarrow \infty}\left| (n^4+3n^3+n+1)/(n^4+3n^3+3n^2+2n) \right|=1\]

So, our expression becomes \[-1<(x-4)<1\]
Note, this is between -1 and 1 becuase the limit was an absolute value.
the radius of convergence is half the distance between the two endpoints, which from this point we can tell will be equal to 1

our interval of convergence requires us solving the inequality the rest of the way:
\[-1<(x-4)<1\]\[3<x<5\]

Answer 6

Alright, that's where I was stuck, I couldn't for the life of me resolve getting the x's isolated.