Find the radius of convergence and the interval of convergence of the series $\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$ Series is:
$$\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$$
So, I understand that I use the ratio test to find r, but I can't simplify the equation to the point where I can do this. Here's where I am so far:
$$
\frac{(n+1)(x-4)^{n+1}}{(n+1)^3+1}\bullet\frac{n^3+1}{n(x-4)^n}
$$
which gets me to
$$
\frac{(n+1)(x-4)(n^3+1)}{n[(n+1)^3+1]}
$$
Which is where I'm stuck. How can I simplify this further so I can find R? Do I just need to multiply this out and look at the coefficients of the highest powers of n?
 A: $$
\lim_{n \to \infty} \left|\frac{(n+1)(x-4)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right|\lim_{n \to \infty} \left|\frac{(n+1)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right| < 1 \\
R = 1 \ \text{and} \ x \in (3, 5)
$$
A: HINT: use the ratio test to find r.
We put
\begin{equation}
a_n=\frac{n}{n^3+1}
\end{equation}
and
\begin{equation}
\lim =\frac{a_{n+1}}{a_n}=1
\end{equation}
Then the series converges for $|x-4|<1$. If $|x-4|=1$, we have
\begin{equation}
\sum_{n=1}^\infty \frac{n}{n^3+1}
\end{equation}
and
\begin{equation}
\frac{n}{n^3+1}\sim \frac{1}{n^2}
\end{equation}
Then
\begin{equation}
\sum_{n=1}^\infty \frac{n}{n^3+1}
\end{equation}
converges, and the original function series converges for $|x-4|\leq 1$.
A: In series convergance what matters is the sequence not the $(x-4)^n$ term. You sequence is $a_n=\frac {n}{n^3+1}$. DO the ratio test and find it.
The centre only plays it's part to find the interval of convergance.For example if the ratio is $1$ what will be the interval with radius $1$ and centre $4$?
