Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n$

share|improve this question
    
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
add comment

5 Answers

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 $nq^n$.

share|improve this answer
    
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
add comment

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 $\,|q|<1\,$

share|improve this answer
    
very nice. Thanx for help –  Mykolas Nov 1 '12 at 8:23
add comment

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]

share|improve this answer
add comment

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 \[ n|q|^n \le \frac{|q|}\eta\cdot (|q| + \eta)^n
\]

share|improve this answer
add comment

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.

share|improve this answer
    
thankyou for help. –  Mykolas Nov 1 '12 at 8:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.