Wednesday, September 23, 2015

sequences and series - limnrightarrowinfty(an+bn)=0 and limnrightarrowinftycn=L imply limnrightarrowinftyexp(an)cnLexp(bn)=0




Consider the sequences {an}nN<0, {bn}nN>0, {cn}nN>0 and suppose
{lim



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


Answer




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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