$\sum_{n=1}^{\infty}c_{n}\cdot x^{n}$ 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".

  • 1
    $\begingroup$ 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


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 $2<|z|<3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.