# The radius of convergence of $\sum_{n=1}^{\infty}\left(c_{n}+d_{n}\right)\cdot x^{n}$

$$\sum_{n=1}^{\infty}c_{n}\cdot x^{n}$$ 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, 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 $$\left|x-a\right| and diverge if $$\left|x-a\right|>R$$.

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

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

If $$(a_n)$$ is any sequence of scalars there exists $$R\in[0,\infty]$$ such that $$\sum |a_n| z^n$$ converges whenever $$|z| and diverges whenever $$|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| implies the sum converges even though $$2<|z|<3$$.