Isn't $f_n(x)=\frac{{n^2}\ln x}{x^n}$ with $x\geq1$ uniformly convergent by $T$-test?

Question

One Dr. showed to me that the function $f_n(x)=\dfrac{{n^2}\ln x}{x^n}$, $x\geq1$ is not uniformly convergent by $T$-test, but I showed it to converge to $0$ anyway. $$\lim_{x\to \infty}T_n=\lim_{x\to \infty} \left[\sup_{x\geq1}\left|f_n(x)-f(x)\right|\right]=\dfrac{3}{4e^{2n}}=0$$ and hence convergent by $T$-test. Thus either of us has to be wrong, please which of us is wrong? Am really waiting.

Answer 1

I think that you are wrong.

Write $f_n(x) = \dfrac{n^2\ln x}{x^n}$, $x\geq 1$. It is clear that $f_n(x) \to 0$ for every $n$, but the convergence is not uniform: we have $$ \sup_{x\geq 1}|f_n(x)-0| \geq f_n(e^{1/n}) = \frac{n}{e} \to +\infty $$

(Notice that this was not a lucky guess: $x=e^{1/n}$ cancels $f_n'(x)$)

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Exercise. Are the following sequences of functions uniformly convergent on $[1,\infty)$? $$ g_n(x) = \dfrac{n\ln x}{x^n}, \qquad h_n(x) = \dfrac{\ln x}{x^n} $$