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Show that $\displaystyle\lim_{n\to\infty} nx^n=0$ for $x\in[0,1)$.

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  • $\begingroup$ @Jay I thought about l'hopitale rule. But this doesn't seem to work. $\endgroup$ Commented Mar 6, 2013 at 21:49
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    $\begingroup$ It looks like it'll work to me. I think you should give it a shot. $\endgroup$ Commented Mar 6, 2013 at 21:50
  • $\begingroup$ @AntonioVargas Should I rewrite it as $\frac{x^n}{1/n}$? And then you get $\frac{nx^{n-1}}{0}$ ? $\endgroup$ Commented Mar 6, 2013 at 21:54
  • $\begingroup$ Wasn't this already asked several times on the site? $\endgroup$
    – Did
    Commented Mar 6, 2013 at 21:55
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    $\begingroup$ Dividing by $0$ means something has gone awry. $\endgroup$ Commented Mar 6, 2013 at 22:04

4 Answers 4

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$$|x|<1:\quad\sum_{n\geq 1} nx^n=\frac{x}{(x-1)^2}$$

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    $\begingroup$ + 10 for this one $\endgroup$ Commented Mar 6, 2013 at 21:58
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Here's the L'Hopital approach. Suppose $x \neq 0$.

$$ \lim_{n \to \infty} n x^n = \lim_{n\to\infty} \frac{n}{x^{-n}} \stackrel{L'H}{=} \lim_{n \to \infty} \frac{1}{- x^{-n} \log x} = -\frac{1}{\log x} \lim_{n \to \infty} x^n = 0. $$

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  • $\begingroup$ I like this one as well ! $\endgroup$ Commented Mar 6, 2013 at 22:09
  • $\begingroup$ But isn't it a problem that this is now a sequence $(f_n)$. And with l'hopitale you normally have the function $f(x)$. I don't know of taking the derivative of sequences. $\endgroup$ Commented Mar 6, 2013 at 22:10
  • $\begingroup$ @90intuition if it makes you more comfortable, replace $x$ with a different letter, say $a$, and $n$ with $x$. Your limit is equal to $$\lim_{x \to \infty} x a^x$$ for $a \in [0,1)$. $\endgroup$ Commented Mar 6, 2013 at 22:28
  • $\begingroup$ In general, when you apply L'Hopital's rule you take derivatives with respect to whichever variable is involved in the limit. You didn't mention that $n$ must be an integer, and both $x^{-n}$ and $n$ are differentiable functions of $n$ if you allow $n$ to take real values, so you can apply L'Hopital's rule by differentiating with respect to $n$. (All of this is referring to the original statement of the problem.) $\endgroup$ Commented Mar 6, 2013 at 22:35
  • $\begingroup$ Even if it was mentioned that $n$ must be an integer, this would still work. In general if you have a sequence $f:\mathbb{N}\to\mathbb{R}$, that may be extended to a differentiable function, we then have all the methods of calculus available to us. $\endgroup$ Commented Mar 7, 2013 at 5:30
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Try to represent x as 1/(1+z) and then from the Bernoulli inequality you have $$(1 + z)^n > 1 + nz $$.

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For $x=0$ it is trivial. For $0\lt x\lt 1$, set $x=\frac{1}{1+t}$. Note that $t$ is positive.

We have $$(1+t)^n \gt \frac{n(n-1)}{2}t^2$$ if $n \ge 2$ (Binomial Theorem). And $n-1\ge n/2$. So $(1+t)^n\gt \frac{n^2}{4}t^2$.

Now if you are looking for an $\epsilon$-$N$ proof you have the ingredients.

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