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$\sum_{n=1}^{\infty}c_{n}\cdot x^{n}$ has radius of convergence 2 and

$\sum_{n=1}^{\infty}d_{n}\cdot x^{n}$ has radius of convergence 3.

then What is the radius of convergence of series $\sum_{n=1}^{\infty}\left(c_{n}+d_{n}\right)\cdot x^{n}$?

Actually I know that

In $\left|x\right|<2$, the series converges. In $2<x<3$, the series diverges.

But I don't know how can show the series diverges in $\left|x\right|>3$.

Can the sum of the two series which are divergent converge or diverge? so how can we know that in $\left|x\right|>3$ that series diverges?

Can we show that by power series theorem? which says that there are only three possibilities: (i) R=0, (ii) = R=$\infty$, (iii) There is a positive number R such that the series converges if $\left|x-a\right|<R$ and diverge if $\left|x-a\right|>R$.

My thought is "That series has R=2 so by power series theorem can't have the other interval".

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    $\begingroup$ If you know it converges for $|z|<2$ and diverges for $2<|z|<3$ that says the radius of convergence is $R=2$. $\endgroup$ Jan 29 at 15:50
  • $\begingroup$ @DavidC.Ullrich The series diverges for $\left|x\right|>3$ is because we have only one R? $\endgroup$ Jan 29 at 15:56
  • $\begingroup$ notice that $\sum_{n=1}^\infty (c_n + d_n)x^n$ = $\sum_{n=1}^\infty c_n x^n$ + $\sum_{n=1}^\infty d_n x^n$ $\endgroup$
    – invictus
    Jan 29 at 16:58

1 Answer 1

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This is clear from the definition of the radius of convergence:

If $(a_n)$ is any sequence of scalars there exists $R\in[0,\infty]$ such that $\sum |a_n| z^n$ converges whenever $|z|<R$ and diverges whenever $|z|>R$.

Cor. If the sum converges for $|z|<2$ and diverges for $2<|z|<3$ then $R=3$.

Because convergence for $|z|<2$ implies $R\ge 2$. But if $R>2$, choose $z$ with $2<|z|<\min(R,3)$; now $|z|<R$ implies the sum converges even though $2<|z|<3$.

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