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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 $r=1$

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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 $1$.

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  • $\begingroup$ Thank you so much. For the radius of convergence, is that true please? (I edited my question). $\endgroup$ – Dima Oct 31 '18 at 19:38
  • $\begingroup$ Yes, your argument is correct $\endgroup$ – José Carlos Santos Oct 31 '18 at 19:55
  • $\begingroup$ Thank you so much. $\endgroup$ – Dima Oct 31 '18 at 19:58

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