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Let $$f_n(t) = \left\{ \begin{array} {lr} t^k e^{-t} \left( 1 - \frac {e^{-t}} n \right)^{n-1}, & t > -\log n \\ 0, & t \le -\log n \end{array} \right.$$ for $k=1,2$. Is there an integrable function which dominates the sequence in the sense that $|f_n(t)| < g$ and $g$ is integrable? For $t > 0$ the sequence is dominated by $t^k e^{-t}$. First of all we have that $f_n(t) \to t^k e^{-t} e^{-e^{-t}}$ and the right hand side is integrable (it is the Gumbel distribution density function). Moreover we have that $f_n(t) * (1 - \frac{e^{-t}}{n})$ is dominated by $ t^k e^{-t} e^{-e^{-t}}$ (in this case the convergence is monotone). I hope $f_n$ is also dominated by some function. Any ideas ?

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I have a new idea:

I found the inequality $1+x \le e^x$ so we have $$( 1 + \frac{-e^{-t}}{n})^{n-1} \le e^{- \frac{n-1}{n} e^{-t}} \le e^{-\frac{1}{2}e^{-t}}$$ and maybe $t^k e^{-t} e^{-\frac{1}{2}e^{-t}}$ is integrable.

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  • $\begingroup$ What is the domain? And what is that weird thing with $\chi$? $\endgroup$ – zhw. Jan 8 '16 at 18:52
  • $\begingroup$ I haven't thought about it and I'm not going to. But I'll say this: Sometimes you can show $\int f_n\to\int f$ by using DCT on $\int_A$ and using MCT on $\int_B$. $\endgroup$ – David C. Ullrich Jan 8 '16 at 18:54
  • $\begingroup$ $f_n(t) = t^k e^{-t} (1 - \frac{e^{-t}}{n})^{n-1}$ for $t > - \log(n)$ and $0$ elsewhere. $\endgroup$ – crankk Jan 8 '16 at 18:54
  • $\begingroup$ The convergence is not monotone for $t<0$ $\endgroup$ – crankk Jan 8 '16 at 19:00

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