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.

  • 1
    $\begingroup$ 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

Let $f_n(x)=\dfrac{\ln(1+nx)}{\ln(1+n)},\,0<x<1.$ $f_n<f_{n+1}\to f(x)=1$. $$a_0^n=0,\,\,a_k^n=\dfrac{(-1)^{k-1}n^k}{k\ln (1+n)}\to \infty(n\to \infty).$$ $a_0=1,a_k=0,k=1,2,\cdots.$

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.