# Radius of Convergence of Sum of two Series.

Hi all, I know there are similar questions on here, but none deal with the fact of trying to prove that $T \geq min\{R,S\}$.
Intuitively this doesn't make sense to me, If you have, $$\sum_{n=n_{0}}^{} (a_{n}+b_{n})x^n$$ Shouldn't this be equal to ;
$$\sum_{n=n_{0}}^{} (a_{n}+b_{n})x^n=\sum_{n=n_{0}}^{} (a_{n})x^n+\sum_{n=n_{0}}^{} (b_{n})x^n$$
And intuitively ...at least for me, shouldn't the radius of convergence of the LHS be the minimum of the radii of convergence of the two power series on the RHS?
I can't think of a case for when it is larger than the minimum of the radii of convergence of the two power series on the RHS.
Any help would be much appreciated.

• As a hint: If $R\neq S$, then you have $T = \min \{R,S\}$. Only if $R = S$ can strict inequality occur. Does that help seeing how it may occur? – Daniel Fischer Apr 21 '14 at 10:39
• A trivial example is taking $a_n = 1$ and $b_n = -1$. Both have radius of convergence $1$ but the sum has radius of convergence $\infty$. – user88595 Apr 21 '14 at 10:40

## 2 Answers

You are correct: as long as both series converge (that is, as long as $\lvert x\rvert\leq\min\{R,S\}$, so that you are inside both radii of convergence), you have $$\sum_{n=n_0}^{\infty}(a_n+b_n)x^n=\sum_{n=n_0}^{\infty}a_nx^n+\sum_{n=n_0}^{\infty}b_nx^n.$$ But, this doesn't mean that $\min\{R,S\}$ is the best that we can do!

For an example of that, think about the case where there is a lot of cancellation of terms in $a_n+b_n$. Do you see what I'm getting at?

But you already know the answer to your question: let $$(a_n)$$ have radius of convergence $$1$$ and $$(b_n)$$ have radius of convergence $$1/2$$. Certainly then, putting $$(c)=(a)+(b)$$, the new $$(c)$$ will have radius of convergence $$1/2$$.

Now what about $$(c)+(-b)$$? Both these series have radius of convergence $$1/2$$, but their sum is $$(a)$$.