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I think should be something very simple... $$\lim_{n\to\infty}nq^{n}$$ wher $|q|<1$ i tried to use binomial theorem, but no success... I know that $q^{n}$ is convergent to $0$, but can't deal with this $n$

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As a worshipper of simplicity, I suggest that my answer may well be the simplest. (See below.) – Michael Hardy Oct 31 '12 at 21:27
up vote 5 down vote accepted

Hint: It is enough to deal with positive $q$. Let $q=e^{-k}$. We are interested in $$\lim_{x\to\infty} \frac{x}{e^{kx}}.$$ Now we can use L'Hospital's Rule.

If you want to use less machinery, let $q=\frac{1}{1+a}$. Use the Binomial Theorem to conclude that if $n\ge 2$ then $$(1+a)^n \ge 1+na+\frac{n(n-1)}{2}a^2\gt \frac{n(n-1)}{2}a^2.$$

Remark: We could simultaneously have dealt with negative $q$, by using absolute values appropriately. That would make things look more complicated than they are. For completeness, note that the case $q=0$ is trivial. And for negative $q$, we have $|nq^n|=n|q|^n$, so since $n|q|^n$ has limit $0$, so does $nq^n$.

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L'Hopital's rule gives correct answers very efficiently, but usually gives little or no insight. – Michael Hardy Oct 31 '12 at 17:40
@MichaelHardy: Agreed! Here it all comes down to exponential kills polynomial. Since saying that would probably not be considered enough, I thought it useful to mention two approaches, one mechanical and the other $\epsilon$-$N$ ready. – André Nicolas Oct 31 '12 at 18:04

Something fancy: the series

$$\sum_{n=0}^\infty x^n$$

has convergence radius equal to $\,1\,$ , so we can derive elementwise the above series and get a series with a convergence radius at least $\,1\,$:

$$\sum_{n-1}^\infty nx^{n-1}$$

Thus, for any $\,|q|<1\,$ , the series

$$\sum_{n=1}^\infty nq^n$$

converges and then $\,nq^n\xrightarrow [n\to\infty]{} 0\,$ for $\,|q|<1\,$

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very nice. Thanx for help – Mykolas Nov 1 '12 at 8:23

Here's another suggestion. Look at the ratio between successive terms, which is $$r_n=\left(1+\frac 1 n\right)q$$ If you look at the eventual behaviour of this from some suitable $N$, so that for $n\geq N, \left(1+\frac 1 n\right)<\frac 1 q$, then $r_n<r_N<1$ and for $k>0$ we have $$(N+k)q^{N+k} \leq r_N^kNq^N (=A_Nr_N^k \text { with constant } A_N)$$ and you can use the result you already have.

[The note in the answer given by André Nicolas about assuming positive $q$ is necessary to make this go through]

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Choose $\eta > 0$ with $|q| + \eta < 1$, then we have \begin{align*} (|q| + \eta)^n &= \sum_{k=0}^n\binom nk |q|^k\eta^{n-k}\\ &\ge n|q|^{n-1}\eta \end{align*} So we have \[ n|q|^n \le \frac{|q|}\eta\cdot (|q| + \eta)^n

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For example, suppose $q = 9/10$. Let $n$ get incremented from $1000$ to $1001$. Then $nq^n$ gets multiplied by $\dfrac{1001}{1000}=1.001$, which is barely more than $1$, and then by $9/10$. You increase it by one-tenth of one percent, then decrease it by ten percent. So it shrinks. And beyond $1001$ it shrinks even faster. So it approaches $0$.

L'Hopital's rule will tell you that it approaches $0$, but the reasoning above shows you why you should expect it to approach $0$, and enables you to feel it.

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thankyou for help. – Mykolas Nov 1 '12 at 8:20

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