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consider a sequence of functions $f_n:(0,\infty)\rightarrow\mathbb{R}$ which are positive and monotone, i.e.

$$0< f_1\leq f_2\leq....\leq f_n\leq f_{n+1}...$$

Now let us assume we know the asymptotic power series for any $f_n$, i.e. we know coefficients $a_k^n\in\mathbb{R}$ with $f_{n}(t)\sim \sum\limits_{k=0}^{\infty}a_k^n t^k$ (as $t\downarrow 0$).

Now let $f:(0,\infty)\rightarrow\mathbb{R}$ be a function with $f(t):=\lim\limits_{n\rightarrow\infty}f_n(x)$ and $f(t)\sim \sum\limits_{k=0}^{\infty}a_k t^k$. Is it maybe possible to show that $\lim\limits_{n\rightarrow\infty}a_k^{n}=a_k$?

I'm very curious about the answer, but I could not find it in any of my books. Neither I could prove or disprove it. Does someone know some fancy examples which can be adapted to this framework?

Best wishes

ps: Initially I had two questions. One of them was answered by Julián Aguirre below, that one can not in general deduce $a_{k}^n\leq a_k^{n+1}$, $\forall k\geq 0$. I changed the question a little bit so that the remaining open question is better visible.

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    $\begingroup$ You can use Dini's theorem to show uniform convergence. This allows you to swap the llimits. This may be useful to show the second statement. $\endgroup$ – Zardo Jun 16 '15 at 9:33
  • $\begingroup$ This is very nice, thank you! At the moment I'm not sure if this is enough to show the convergence of the coefficients. I need to think about it for a time. $\endgroup$ – asd Jun 16 '15 at 13:18
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Let $f_n(x)=\dfrac{\ln(1+nx)}{\ln(1+n)},\,0<x<1.$ $f_n<f_{n+1}\to f(x)=1$. $$a_0^n=0,\,\,a_k^n=\dfrac{(-1)^{k-1}n^k}{k\ln (1+n)}\to \infty(n\to \infty).$$ $a_0=1,a_k=0,k=1,2,\cdots.$

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  • $\begingroup$ Welcome to Math.SE! Can you try to explain the steps you are taking. Right now, your answer is kind of difficult to read. $\endgroup$ – Hrodelbert Jun 17 '15 at 13:25
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Since you are interested in the asymptotic behaviour as $t\to0$, my example will be defined in $(0,1)$. Let $f_n(t)=1-t^n$, $t\in(0,1)$. Then $0<f_n(t)<f_{n+1}(t)$ fot all $t\in(0,1)$. On the other hand $$ a_0^n=1,\quad a_k^n=\begin{cases}0 & k\ne ,\\ 1 & k=n.\end{cases} $$ For no $k>1$ is $a_k^n$ increasing.

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  • $\begingroup$ Thank you for this very helpful answer! It helped a lot. $\endgroup$ – asd Jun 16 '15 at 13:07

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