Let $$f_{n}(t) = \frac{n^2t^2}{(1+t^2)^{n}}$$

I have to count limits

  1. $$ \lim_{n \to +\infty}(\lim_{t \to 0} f_{n}(t))$$
  2. $$ \lim_{t \to 0}(\lim_{n \to +\infty} f_{n}(t))$$
  3. $$\lim_{n \to +\infty}f_{n}(\frac{1}{n})$$

My results: 1st and 2nd are equal to 0, 3rd is equal to 1. My question, is this sequence uniformly convergnent for all real values of t? I think it won't be because of this $\frac{1}{n}$ points, but how to write it formally?


Hint. One may consider $$ t \mapsto f_n(t):=\frac{n^2t^2}{(1+t^2)^{n}},\quad t \in \mathbb{R}. $$ By differentiating $f_n$ one gets $$ f'_n(t)=-2n^2t\:\frac{(n-1)t^2-1}{(1+t^2)^{n+1}} $$ and, as $n \to \infty$, $$ \sup_{t\in \mathbb{R}}\left|f_n(t) \right|=f'_n\left(\pm \frac1{\sqrt{n-1}}\right)=\frac{(n-1)^{n-1}}{n^{n-2}} \sim \frac{n}e \to +\infty. $$ The pointwise convergence is to $0$, thus the convergence can't be uniform on $\mathbb{R}$.

  • $\begingroup$ But can we say that our sequence is not uniformly convergent without using differentation? $\endgroup$
    – janusz
    Apr 19 '16 at 16:47

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