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Intuitively it's easy, but hard to prove by the epsilon-delta method:

$$ \lim_{n \to \infty} n x^{n} = 0$$

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Well, here again I'll try a fancy proof. Let us look at the power series

$$\sum_{n=1}^\infty nx^n\,\,,\,\,\text{and let us define}\,\,\,a_n:=nx^n$$

We try the ratio test to find this series convergence radius:

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(n+1)x^{n+1}}{nx^n}\right|=|x|\frac{n+1}{n}\xrightarrow [n\to\infty]{} |x|$$

Thus, the series converges (absolutely, even) for $\,|x|<1\,$ , from which it follows that the series general term must converge to zero, i.e.

$$a_n=nx^n\xrightarrow [n\to\infty]{} 0\,\,,\,\,\text{for}\;\;|x|<1$$

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Nice proof Don Antonio ; however, if we use d'Alembert criterion for power series, shouldn't we set $a_n=n$ ? We then get $lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}=\frac{n+1}{n}=1$ ; and then we find that the radius of convergence of $\sum_{0}^{\infty}nx^n$ is 1, so that the series converges (absolutely) on $]0,1[$ ; therefore $lim_{n\to\infty} \space nx^n$ must be $0$ if $|x|<1$ ? – Asinus Dec 9 '13 at 11:30
Doing what you propose, @Asinus, will give us the convergence radius, which indeed is $\;1\;$ , but I prefer many times (perhaps ever most times) to take the whole thing as it gives us directly the conditions on the variable for convergence. – DonAntonio Dec 9 '13 at 13:05
I didn't know we could do that. It's pretty cool. – Asinus Dec 9 '13 at 15:45

We prove the result under the slightly weaker condition $|x|\lt 1$.

Let $|x|=\dfrac{1}{1+t}$. Then $t\gt 0$.

By the Binomial Theorem, if $n \ge 2$, then $$(1+t)^n \ge 1+nt +\frac{n(n-1)}{2}t^2 \gt \frac{n(n-1)}{2}t^2.$$

It follows that $$0\le n|x^n| \lt \frac{2}{(n-1)t^2}.$$ Now it is easy, given $\epsilon \gt 0$, to find $N$ such that if $n \gt N$ then $\dfrac{2}{(n-1)t^2}\lt \epsilon$.

Remark: If we do not wish to use the Binomial Theorem, let $m=\left\lfloor\frac{n}{2}\right\rfloor$. By the Bernoulli Inequality, $(1+t)^m \ge 1+mt$, and therefore $(1+t)^n \ge (1+mt)^2\ge m^2 t^2$.

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Suppose you know $\lim n^{\frac{1}{n}} =1$, let $\epsilon >0$

then take $n_o$ such that $$n^{\frac{1}{n}} \leq 1+\frac{a-x}{x} \forall n \geq n_0$$ where $$x<a<1$$ Now assume you know $\lim b^n =0$ for $b\in (0,1)$ Choose a $n_1$ such that $$a^n \leq \epsilon \forall n \geq n_1$$ Now take $ \max\{n_0,n_1\}$ so we get $$nx^n= (n^{\frac{1}{n}}x)^n \leq ((1+\frac{a-x}{x})x)^n=a^n\leq \epsilon$$

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Let $a_n=nx^n$. You have $$\frac{a_{n+1}}{a_n}= \frac{n+1}n x = \left(1+\frac1n\right) x.$$ Let $q=\frac{1+x}2$. (Or we could choose any $q$ such that $x<q<1$.)

There exists $n_0$ such that $\left(1+\frac1n\right) x \le q$ for $n\ge n_0$. (Since $\lim\limits_{n\to\infty} \left(1+\frac1n\right) x = x < q$.)

Thus we have $$0 \le a_n \le a_{n_0} \cdot q^{n-n_0}$$ for $n\ge n_0$. (You can show this easily by induction.)

Since $q^n\to 0$ for $n\to\infty$ we get $$\lim_{n\to\infty} a_n=0.$$

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We have $nx^n=\exp(\log n)\exp(n\log x)=\exp(n\log x+\log n)$, so it's enough to show that $n\log x+\log n\to -\infty$ as $n\to +\infty$. We use the fact that $\log n\leq \sqrt n$ for $n$ large enough to see that $$n\log x+\log n\leq n\log x+\sqrt n=n\left(\log x+\frac 1{\sqrt n}\right).$$ As $\log x<0$, $\log x+\frac 1{\sqrt n}<\frac{\log x}2$ for $n$ large enough hence $$n\log x+\log n\leq n\frac{\log x}2,$$ which gives the result.

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Isn't it enough to know that the exponential function with $a>1$ goes faster to infinity than the polynomial function?

$$\lim_{n\to\infty}\frac{n}{a^n}=\lim_{n\to\infty}\frac{e^{\ln n}}{e^{n \ln a}}=0.$$ Chris.

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I would propose another approach. Notice that the sequence $$b_n:=\frac{1}{x^n}$$ is positive, strictly increasing and unbounded.

Define $$a_n:=n$$ as well.

Stolz Cesaro theorem is applicable and therefore, if $$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ exists and it is equal to $l\in\mathbb R$, then also the limit proposed is equal to $l$.

Then we are reduced to evaluate $$\lim_{n\to\infty}\frac{1}{\frac{1}{x^{n+1}}-\frac{1}{x^n}}=\lim_{n\to\infty}\frac{x^{n+1}}{1-x}=0.$$ To tackle the last equality the $\varepsilon$-$\delta$ argument is perfectly fine and easy to perform. To conclude, this argument finishes the proof.

In a way you have not cheated too much because $\varepsilon$-$\delta$ is used. Hope you liked it. Mathematicians are lazy :P.


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lol, I think you're right! :-) – user 1618033 Sep 20 '12 at 16:45

This will follow from a more general result. If $\{a_n\}\geq 0$ be a monotone decreasing sequence such that $\sum a_n <\infty$ then $na_n\rightarrow 0. $

Let $\epsilon >0$ be arbitrary . Then there exists a natural number $m$ such that for $p\in \mathbb N $
$$a_{m+1}+a_{m+2}+..+a_{m+p}<\epsilon $$ But $\{a_n\}$ is monotone decreasing means $$pa_{m+p}\leq a_{m+1}+a_{m+2}+..+a_{m+p }< \epsilon$$ putting p-=m we get $$2ma_{2m}< 2\epsilon$$ Also putting $p=m+1$ we get $$(m+1)a_{2m+1}<\epsilon$$ Hence $$(2m+1)a_{2m+1}\leq (2m+2)a_{2m+1}<2\epsilon$$ Hence both the sequences $\{2ma_{2m}\}$ and $ \{(2m+1)a_{2m+1}\}$ converges to $0$. Hence the result follows.

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