1
$\begingroup$

enter image description here

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Daniel Fischer Apr 21 '14 at 10:39
  • 1
    $\begingroup$ 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 $\infty$. $\endgroup$ – user88595 Apr 21 '14 at 10:40
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $(a)$.

$\endgroup$

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.