Find a power series in $x$ that has the given sum, and specify the radius of convergence.
$\frac{x^2+1}{x-1}$
$\frac{x^2+1}{x-1}=-(x^2+1)\frac{1}{1-x}$
$=-1-x-2 \sum_{2}^{\infty} x^n$
Now, is the radius of convergence will be changed if I delete the first two terms of the series?
I know that, delete the first 2 terms of the series has no effect on its convergence.
Since I can’t found the form of n-th term of the series with out delete the first two terms.
$lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|= lim_{n \to \infty} |\frac{x^{n+1}}{x^n}|=|x|$.
By the ratio test, the series is convergent if $|x|<1$, so the radius of convergence is $r=1$
No, deleting the first two terms (or any finite amount of terms, for that matter) will not change the redius of convergence.
By the way, your answer is correct:$$\frac{x^2+1}{x-1}=-1-x-2\sum_{n=2}^\infty x^n.$$The radius of convergence of this power series is $1$.