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Consider the sequences $\{a_n\}_{\forall n \in \mathbb{N}}<0$, $\{b_n\}_{\forall n \in \mathbb{N}}>0$, $\{c_n\}_{\forall n \in \mathbb{N}}>0$ and suppose $$ \begin{cases} \lim_{n\rightarrow \infty} (a_n+b_n)=0\\ \lim_{n\rightarrow \infty} c_n=L<\infty \end{cases} $$

Could you help me to show that $$ \lim_{n\rightarrow \infty} [\exp(a_n)*c_n-L*\exp(-b_n)]=0 $$ ?


I know that by assumption $$ \lim_{n\rightarrow \infty} [\exp(a_n)*c_n-L*\exp(-b_n)]= \lim_{n\rightarrow \infty} [\exp(-b_n+o(1))*(L+o(1))-L*\exp(-b_n)] $$ where $o(1)$ is a number going to zero as $n\rightarrow \infty$. How can I proceed from here?


Let me add another assumption (thanks to a comment below) $$ \exp(a_n)\equiv \Pi_{k=1}^{2n} x_{n,k} $$ where $x_{n,k}\in [0,1]$ and $\lim_{n\rightarrow \infty} x_{n,k}=1$ $\forall k$

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  • $\begingroup$ Do we have any assumption on $\{a_n\}$ other than negative ? is it increasing (so it has a limit) ? If $\{a_n\}$ has a limit, you just need to factorize by $\exp(a_n)$ and you're done $\endgroup$ – Jean Rostan May 11 '18 at 17:01
  • $\begingroup$ I have that $a_n\equiv \frac{1}{n}\sum_{k=1}^{2n} n \log(x_{k,n})$ so that $\exp(a_n)=\Pi_{k=1}^{2n}x_{k,n}$. Moreover, $\lim_{n\rightarrow \infty}x_{k,n}=1$ and $x_{k,n}\in (0,1)$ $\forall k$. $\endgroup$ – TEX May 11 '18 at 17:12
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You can just compute $$ |e^{a_n}c_n - L e^{-b_n}| = e^{a_n}|c_n - L e^{-(a_n+b_n)}| \leq |c_n - L e^{-(a_n+b_n)}|. $$ Now $c_n\to L$ and $a_n+b_n\to 0$. So, because $e^x$ and $|x|$ are continuous functions, you can "pass to the limit inside the functions" and get $$ |e^{a_n}c_n - L e^{-b_n}| \leq |c_n - L e^{-(a_n+b_n)}| \to |L-Le^{-0}| = 0. $$ By the comparison principle, you get convergence of that guy to 0.

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In these kind of problems, I try to get the terms that go to zero by themselves. As long as the other terms behave (usually meaning that they are bounded), the rest usually follows.

$\begin{array}\\ \exp(a_n)*c_n-L*\exp(-b_n) &=\exp(a_n)*(c_n-L+L)-L*\exp(-a_n-b_n+a_n)\\ &=\exp(a_n)*(c_n-L)+\exp(a_n)L-L*\exp(-a_n-b_n)\exp(a_n)\\ &=\exp(a_n)*(c_n-L)+L\exp(a_n)\left(1-\exp(-a_n-b_n)\right)\\ \end{array} $

Combine this with your assumptions that $c_n-L \to 0$ and $a_n+b_n \to 0$ and you are done.

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  • $\begingroup$ But one has zero assumptions on $\exp{a_n}$ on its convergence, so can't conclude from it. $\endgroup$ – Jean Rostan May 11 '18 at 20:30

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