# Prove/disprove Radius of convergence of Power series

Let $\sum_{n=0}^\infty a_n x^n$ and $\sum_{n=0}^\infty b_n x^n$ Be two power series with radius of convergence of $R_1$ and $R_2$ respectively and let $R$ be the radius of convergence of the series $\sum_{n=0}^\infty (na_n - b_n) x^n$

Then $R$ is at least $\min\{R_1,R_2\}$

I think I should prove it (after many tries of disproving).

The question is how should I approach this?

Thank you

## 5 Answers

hint

If $\sum a_nx^n$ has $R_1$ as radius of convergence, then $\sum na_nx^n$ has the same radius $R_1$ .

use the sum of two series.

There are theorems asserting that

• a power series $\sum_{n=0}^\infty a_n x^n$ is differentiable at every point in the interior of its region of convergence (althouth not necessarily on the boundary), and it can be differentiated term by term, yielding $\sum_{n=0}^\infty na_n x^{n-1},$ and
• the radius of convergence of the power series for the derivative is the same as that of the series you started with.

Thus $x\sum_{n=0}^\infty na_n x^{n-1}$ has radius of convergence $R_1.$

The series $\sum_{n=0}^\infty (na_n-b_n) x^n = x\sum_{n=0}^\infty a_nx^{n-1} - \sum_{n=0}^\infty b_n x^n$ converges at every value of $x$ at which the original two series converge, with the possible exception of values of $x$ that are on the boundary of one of the regions of convergence.

Remember that $$\frac{1}{R_1} = \limsup \sqrt[n]{a_n}$$ and notice that $$\sqrt[n]{|n a_n - b_n|} \le \sqrt[n]{n|a_n| + |b_n|} \le \sqrt[n]{n}\sqrt[n]{|a_n|+|b_n|}.$$ Now suppose that $\limsup \sqrt[n]{|a_n|} \ge \limsup \sqrt[n]{|b_n|}$, you find $$\limsup \sqrt[n]{|a_n| + |b_n|} = \limsup \sqrt[n]{|a_n|}\sqrt[n]{1 + \frac{|b_n|}{|a_n|}} = \limsup \sqrt[n]{|a_n|}.$$

• Suppose that there exist an infinite number of $n$ such that $a_n=0$. As you divide by $a_n$, I think that there is a problem. – Kelenner Jun 25 '17 at 15:54

Since the radius of convergence of $\sum\limits_{n=0}^\infty na_nx^n$ is given by $$\frac1{\limsup\limits_{n\to\infty}|na_n|^{1/n}}=\lim_{n\to\infty}\frac1{n^{1/n}}\frac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}}=R_1$$

If $R_1\ne R_2$, then the radius of convergence of $$\sum_{n=0}^\infty(na_n-b_n)x^n$$ is $\min(R_1,R_2)$. However, if $R_1=R_2$, the radius of convergence of the series could be larger than $\min(R_1,R_2)$. Consider $a_n=1$ and $b_n=n$.

The same result as indicated by Emanuel Paolini can be reached via application of the ratio test plus triangle inequality followed by some basic algebra.