According to Rudin Since $\lim_{n\rightarrow\infty}n^{\frac{1}{n}}=\limsup_{n\rightarrow\infty}n^{\frac{1}{n}}=1$ and $\limsup_{n\rightarrow \infty}|c_n|^{\frac{1}{n}}=R$, $$\limsup_{n\rightarrow \infty}(n|c_n|)^{\frac{1}{n}}=\limsup_{n\rightarrow \infty}|c_n|^{\frac{1}{n}}=R.$$ Why is this reasonable?
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Are we have any theorem such that $limsup_{n\rightarrow\infty}a_n=a,limsup_{n\rightarrow\infty}b_n=b,\ then\ limsup_{n\rightarrow\infty}a_nb_n=ab $? – 백주상 Aug 10 '17 at 05:24
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You are using here: If $\lim x_n=A$ and $\limsup y_n=B$ where $A,B>0$, then $\limsup x_ny_n=AB$. (In fact, I think $A>0$ suffices here.) You can have a look here: If $\lim_{n\to \infty}a_n = a\in \mathbb{R}$ . Prove that $\limsup_{n\to \infty}a_n x_n=a\limsup_{n\to \infty}x_n$. (And you can probably find a few other similar posts.) – Martin Sleziak Aug 10 '17 at 05:27
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I think you at least need to specify the sign (if both sequences are negative, this doesn't always hold). Intuitively, if the sequences are positive, we should have $\lim\sup a_nb_n \leq ab$ (we can't guarantee equality - can you think of any counterexamples?) – platty Aug 10 '17 at 05:29
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Thank you for your answers, now I could search a theorem in Elementary analysis of Kenneth A.Ross.(Theorem 12.1 said that) – 백주상 Aug 10 '17 at 05:59
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Then following fact is used:
Lemma. If $a_n\to a>0$ and $\limsup b_n=b\ge 0$, then $\limsup a_nb_n=ab$.
Proof. It suffices to show the following:
(i) If a subsequence of $a_nb_n$ converges to $c$, then $c\le ab$.
(ii) There exists a subsequence of $a_nb_n$ converging to $ab$.
For (i), if $a_{n_k}b_{n_k}\to c$, then $$ b_{n_k}=\frac{1}{a_{n_k}}\cdot a_{n_k}b_{n_k}\to \frac{1}{a} \cdot c, $$ and hence $\frac{c}{a}\le b$ or $c\le ab$.
Fot (ii), since $\limsup b_n=b$, there exists a subsequence $b_{n_k}\to b$ and hence $a_{n_k}b_{n_k}\to ab$.
Yiorgos S. Smyrlis
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