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Am I right in saying that the sequence of functions $$f_n(x)=\displaystyle\frac{xn^\alpha}{e^{nx}\times\ln(n)}$$ converges pointwise to 0 $\forall{x}\in\mathbb{R}$?

Thanks for any help

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You are right for $x \ge 0$. – André Nicolas Oct 27 '11 at 23:27
up vote 4 down vote accepted

No, take $x=-1$ then you get $\displaystyle \frac{-e^n n^{\alpha}}{\log(n)}\to-\infty$

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oh dear of course, is it for all positive x then? – hmmmm Oct 27 '11 at 23:27
Yes, for $x\geqslant 0$ this is true since $\displaystyle \frac{1}{e^{xn}}$ dominates the rest of the terms. – Alex Youcis Oct 27 '11 at 23:28
cool thanks, that's what I thought just had a bit of a brain lapse. Thanks for the help – hmmmm Oct 27 '11 at 23:30
Then what can we say abou the convergence for $x<0$? That it diverges to $-\infty$? – Jelly Belly Feb 5 at 23:11

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