Let $t_n\ge 0$. How to prove that $\lim\limits_{n\to \infty} \sup t_n=\lim\limits_{n\to \infty} \sup t_n\sqrt[n]{n}$?

Here is my sketch: So $t_n\ge 0$ and $\sqrt[n]{n}\ge 1$ then $t_n\sqrt[n]{n}\ge t_n$ hence $$\lim\limits_{n\to \infty} \sup t_n\sqrt[n]{n}\ge\lim\limits_{n\to \infty} \sup t_n.$$ How to prove the converse inequality?

  • $\begingroup$ Hint: $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$. $\endgroup$ – Wojowu Dec 13 '15 at 15:49
  • $\begingroup$ I know that fact but I can't use it :( $\endgroup$ – Raheem Najib Dec 13 '15 at 15:49
  • $\begingroup$ Taking $t_n=1$ you can notice that the equality of lim sups in your question implies $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$. I don't see any real way to answer your question without proving that limit as an intermediate step. $\endgroup$ – Wojowu Dec 13 '15 at 15:51
  • $\begingroup$ @RaheemNajib what Wojowu mentioned is a very basic fact. Why aren't you allowed to use it? $\endgroup$ – user258700 Dec 13 '15 at 15:53
  • $\begingroup$ I can't understand you. Why you put $t_n=1$? It's meaningless since $t_n$ is arbitrarily sequence. $\endgroup$ – Raheem Najib Dec 13 '15 at 15:53

We can prove more general claim: If $\lim \limits_{n\to \infty}a_n=1$ and $\lim\sup \limits_{n\to \infty}b_n=b$ then $\lim\sup \limits_{n\to \infty}a_nb_n=b.$

Proof: Let $X_1$ be the set of all subsequential limits of sequence $\{b_n\}$ and $X_2$ be the set of all subsequential limits of sequence $\{a_nb_n\}$. We prove that $X_1=X_2$.

Let $s\in X_1$ then $\exists \{n_k\}$ such that $b_{n_k}\to s$ as $k\to \infty$. Then $a_{n_k}b_{n_k}\to s$. Then $s\in X_2$ and $X_1\subseteq X_2$. Making converse reasoning we get that $X_2\subseteq X_1.$ Thus $X_1=X_2$ and $\sup X_1=\sup X_2$. Hence $$\lim\sup \limits_{n\to \infty}a_nb_n=\lim\sup \limits_{n\to \infty}b_n.$$

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