Consider a function sequence $\{f_n(t)\}$ ($n\in\mathbb{N}^+$) defined on the interval $(\frac{1}{2},1)$, where \begin{eqnarray}\label{eqn:constraint1} f_n(t)=\frac{\exp\left(n\left(\log R(h_t) - th_t\right)\right)}{h_t\sigma(h_t)\sqrt{2\pi n}}\left(1+o\left(1\right)\right) \end{eqnarray} as $n\to \infty$ uniformly in $t$ in any closed subinterval of the interval $(\frac{1}{2},1)$, \begin{eqnarray} R(h)=\frac{e^h-1}{h}, \end{eqnarray} \begin{eqnarray} m(h)=\frac{R'(h)}{R(h)}=\frac{e^h}{e^h-1}-\frac{1}{h}, \end{eqnarray} \begin{eqnarray} \sigma(h)=m'(h), \end{eqnarray} and $h_t$ is the positive real root of the equation $m(h)=t$.

Then we study the following equation of $g_n$ \begin{eqnarray} c=g_n+\frac{\int_{g_n}^1f_n(t)~dt}{f_n(g_n)},~\frac{1}{2}< g_n<1, \end{eqnarray} where the constant number $\frac{1}{2}< c<1$.

Try to prove:

  1. the convergence of the sequence $\{g_n\}$ and compute $A=\lim_{n\to \infty}g_n$;
  2. $\lim_{n\to \infty}n(g_n-A)=0$.
  • $\begingroup$ series or sequence? $\endgroup$ – Zhou Aug 17 at 3:56
  • $\begingroup$ Sorry for this typo. I have corrected it. $\endgroup$ – Ryan Chen Aug 17 at 7:13
  • $\begingroup$ did you try to code it up? for $o(1)$ just try different things. $\endgroup$ – Sina Baghal Aug 18 at 5:22
  • $\begingroup$ Can you tell where this problem comes from please. Sometimes a problem is more easily solved in its original form and then follows by applying an appropriate continuous map. $\endgroup$ – Dieter Kadelka Aug 18 at 8:31

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