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..


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

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

  • $\begingroup$ 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

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