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Study the pointwise and uniform convergence of $f_n(t)=n^{n^t}$ for $t\in\mathbb{R}$.

It is $f_n \to 1 $ pointwise if $t<0$, while $f_n \to \infty$ for $t \geq 0$; since $f_n(t)=e^{n^t \log n}$ it is $f_n'(t)=n^t  e^{n^t \log n}\log^2n>0$ so $f_n$ is strictly increasing for all $t\in\mathbb{R}$. Let $b>0$ and consider $t\in(-\infty,-b]$; since $f_n$ is increasing, its supremum is reached for $t=-b$ and so
$$\sup_{t \in (-\infty,-b]} |f_n(t)-1|=\sup_{t\in(-\infty,-b]}\left(e^{n^t \log n}-1\right)=e^{n^{-b} \log n}-1\to0 \ \text{for} \ n \to \infty$$
So $f_n$ is uniformly convergent to $1$ for all $t\in(-\infty,-b]$; is this correct until now?

My question is about the case when I have to consider the interval $[-b,0)$: my lecturer says that
$$\sup_{t \in [-b,0)} |f_n(t)-1|=\sup_{t \in [-b,0)} |e^{n^t \log n}-1|=1$$
But I don't understand why the supremum is $1$ in this case; I would say that it is reached when $t\to0^{-}$ because $f_n$ is increasing, so I would say that the supremum is $n-1$, what am I missing? Thanks.

 Feb 6, 2021
 #1
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Judging by the examples below, you are right and your lecturer is wrong!

 

 Feb 6, 2021
 #2
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Thanks for your answer Alan! Can I ask you what software are you using to plot the $f(t,n)$? Have a good day.

 Feb 6, 2021
 #3
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I used MathCad, but you could probably do the plot in Desmos (https://www.desmos.com/calculator).

Alan  Feb 6, 2021
edited by Alan  Feb 6, 2021
edited by Alan  Feb 6, 2021

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