# Find a power series in $x$ that has the given sum, and specify the radius of convergence.

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 $$r=1$$

## 1 Answer

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 $$1$$.

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