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I'm an undergraduate mathematics student and I'm trying to understand the basics of limes superior. I recently got stuck on the following question:

Let $(a_n)$ be a sequence of numbers and let $(b_n)$ be a sequence of numbers with $\lim_{n\to\infty}b_n=1$. Show:

1) $$\lim\limits \sup (b_na_n)=\lim\limits \sup a_n$$ 2) $$\lim\limits \sup (a_n^{b_n})=\lim\limits \sup a_n$$

My first thought on trying to solve problem 1) was to try and show that $\lim\limits \sup (b_na_n)$ = $\lim\limits \sup (b_n) * \lim\limits \sup (a_n)$, because then I can use the fact that $\lim\limits \sup (b_n) = \lim\limits_{n \rightarrow \infty} b_n = 1$, and that would be sufficient. If it's possible to show it this way, I think the second problem would be solvable in a similar fashion. However I'm not really sure how to show this and I don't even know if I'm thinking in the right direction. Anyone mind helping me out on this one?

durianice
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Spacer
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    The second statement follows from the first upon taking logarithms (and remembering that limits commute with continuous functions). – Greg Martin Jan 08 '14 at 22:33
  • Do you mean I should take the log of $\lim\limits \sup (a_n^{b_n})$ ? – Spacer Jan 09 '14 at 10:45
  • For any positive sequence $c_n$, the lim sup of $\log c_n$ is the same as the logarithm of the lim sup of $c_n$. - Also, you conjectured that $\limsup a_nb_n = \limsup a_n \times \limsup b_n$, and that's a reasonable thought, but it turns out to be wrong. Consider for example $a_n = 1 + (-1)^n$ and $b_n = 1 - (-1)^n$. – Greg Martin Jan 09 '14 at 18:21
  • For positive sequences you have $\liminf a_n \cdot \limsup a_n \le \limsup (a_nb_n) \le \limsup a_n \cdot \limsup b_n$. See this question and references given there: http://math.stackexchange.com/questions/113121/lim-sup-inequality-limsup-a-n-b-n-leq-limsup-a-n-limsup-b-n (The equality does not hold in general, I think it is not very difficult to find counterexamples; you might try some sequences consisting only of 0's and 1's.) – Martin Sleziak Jan 15 '14 at 11:57

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By definition of limit:

$\forall\epsilon>0\ \ \exists n(\epsilon)\in{\Bbb N}\ \ \forall n>n(\epsilon)\ \ 1-\epsilon < b_n < 1+\epsilon$.

Do the same for $\lim\sup a_n$ (http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior) and use the inequalities on $b_n$ and $a_n$.

Martín-Blas Pérez Pinilla
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