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Prove that $$\lim_{n\rightarrow \infty}nx^n = 0$$ for $|x| < 1$.

I can tell that $\lim_{n\rightarrow \infty}x^n = 0$, and it's going to zero much faster than $n$ goes to $\infty$. But how do I take it from here? And how do I also prove it for negative x's?

iTayb
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  • Idea: If $x=0$ there is nothing to do. Otherwise let $|x|=\frac{1}{1+t}$. If $n\ge 2$ then by the binomial theorem $(1+t)^n \ge 1+tn+(n)(n-1)t^2/2\gt n(n-1)t^2/2$. – André Nicolas Jan 14 '16 at 01:33
  • Is it possible to prove this theorem with limit of sequence definition, that is in $\epsilon$ language? – Salech Alhasov Jan 14 '16 at 01:37

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It is a bit round about but you can note that for $|x|<1$ that $\sum\frac{1}{x^n}=\frac{1}{1-x}$, and then it is known that we can differentiate both sides of the expression, and we see that $\sum nx^{n-1}=\frac{1}{(1-x)^2}$, and thus that $\sum nx^n=\frac{x}{(1-x)^2}$ for all $|x| < 1$. Thus since the terms of a convergant sequence must go to 0, you have your result.

siegehalver
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