Does the index in the series affects how I calculate the radius of convergence?

Question

I was given the series $\sum_{n=2}^\infty \frac{(n+4)}{n^2-1} (x-3)^{n+1}$, where I want to find its limit of convergence by using $R:=\frac{1}{\lim sup_{n\to\infty}|c_n|^{(1/n)}}$

However, I am struggling with two problems, first is that the index of the series starts from n=2 instead of n=0, does this affect how my calculation for the radius of convergence?

Secondly, the $(x-3)$ is to the power of $n+1$ instead of $n$. how do I convert it so that it matches the power series form $\sum_{n=0}^\infty c_n (x - a )^n$ ?

Thank you for your answers.

Answer 1

For your first question, $$\lim_{n\to\infty} a_n =\lim_{n\to\infty} a_{n+k}$$ for any $k$.

For your second question, the expression you have above tells you $c_n=0$ for $n\le 2$. Can you see how to get the other $c_n$?

EDIT: For example, $c_3$ is the coefficient of $(x-3)^3$, so

$$c_3=\frac{2+4}{2^2-1}$$

Put

$$d_n:=\frac{n+4}{n^2-1}$$

Then $c_n=d_{n-1}$ whenever $n\ge 3$.

Answer 2

You can set $k = n+1$ and rewrite the series as $$\sum_{n=2}^\infty \frac{(n+4)}{n^2-1} (x-3)^{n+1} = \sum_{k=3}^\infty \frac{k+3}{(k-1)^2-1} (x-3)^k = \sum_{k=3}^\infty \frac{k+3}{k^2-2k} (x-3)^k$$ or $\sum_{k=0}^\infty c_k(x-3)^k$ where $$c_k = \begin{cases} 0, &\text{if } k \le 2\\ \frac{k+3}{k^2-2k}, &\text{if } k \ge 3\end{cases}$$ Therefore the radius of convergence is $$R = \frac1{\limsup_{n\to\infty}\sqrt[k]{|c_k|}} = \frac1{\limsup_{n\to\infty}\sqrt[k]{\frac{k+3}{k^2-2k}}} =1$$ since we can disregard the first three terms of $(c_n)_n$ when taking $\limsup$.