0
$\begingroup$

Let $\sum_{n=0}^{\infty} a_nz^n$ and $\sum_{n=0}^{\infty} b_n z^n$ be two power series with respective radii of convergence $R_1 $ aand $R_2$.Assume that there is a positive constant $M$ such that $|a_n| \leq M |b_n|$ for all but finitely many $n.$ Prove that $R_1 \geq R_2.$

If we consider $R_1={1\over \limsup_{n\to \infty}|a_n|^{1/n}}$ and $R_2={1\over \limsup_{n\to \infty}|b_n|^{1/n}}$ then from given inequality we get $R_2 \leq R_1 \limsup_{n \to \infty}M^{1/n}$. But from here how can we conclude the result?

Please someone give some hints.. Thank you..

$\endgroup$
1
$\begingroup$

$ \lim \sup |a_n|^{1/n} \le \lim \sup M^{1/n}|b_n|^{1/n}= \lim \sup |b_n|^{1/n}$,

since $\lim \sup M^{1/n}= \lim M^{1/n}=1$

$\endgroup$
  • $\begingroup$ Thanks..It didn't come into my mind that $\lim_{n \to \infty} M^{1/n}=1$ $\endgroup$ – nurun nesha Sep 5 '17 at 9:28

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.