Why is it that $\sum na_n x^{n-1}$ and $\sum na_n x^n$ have the same radius of convergence $R$?
How can I show that $$\lim \sup|na_n|^{1/n} = \lim \sup |na_n x|^{1/n}$$ since this would mean that both series have the same radius of convergence $R$?
Asked
Active
Viewed 65 times
1
-
2You can directly note they have the same radius of converge because one is just x times the other – mathworker21 Dec 18 '18 at 03:54
-
That is exactly what I want to prove – The Bosco Dec 18 '18 at 11:51
-
can you be more specific? – mathworker21 Dec 18 '18 at 11:55
1 Answers
0
Instead of using any formula for radius of convergence simply use the fact that $\sum na_nx^{n-1}$ converges iff $\sum na_nx^n$ does, for any $x \neq 0$. Obviously, this implies that they have the same radius of convergence.
Kavi Rama Murthy
- 311,013
-
But proving that part that one converges iff the other one does is what I don't get. – The Bosco Dec 18 '18 at 11:09
-
If $\sum a_n $ converges then $\sum c a_n$ also converges for any constant $c$. Apply this with $c=x$ in one direction and $c =\frac 1 x$ in the other direction. – Kavi Rama Murthy Dec 18 '18 at 11:43