Prove that $\lim_{n\to∞}(nq^n)=0$, while $\lvert q\rvert <1$

Question

Prove that: $$\lim_{n\to∞}(nq^n)=0$$ ($\lvert q\rvert <1$)

My only thought on this one is to treat q as a sub-series: $q=$ $1 \over b$ $\Rightarrow$ $q^n=$ $1\over b^n$ ;

So I can rewrite this as $$\lim_{n\to∞}({n \over b^n})=0$$ Any clues on how to solve this one?

I have to show that there is an $N \in \Bbb N$, and $\varepsilon >0$ which for every $n>N$ the following is true: $|nq^n|<\varepsilon$

please see this http://math.stackexchange.com/questions/197522/how-to-prove-that-lim-n-to-infty-n-xn-0-when-0x1/1515009#1515009 — Black-horse, Nov 14 '15 at 14:28
Another $\epsilon$ answer has been given here:http://math.stackexchange.com/questions/197522/how-to-prove-that-lim-n-to-infty-n-xn-0-when-0x1/197612#197612 — Olivier Oloa, Nov 14 '15 at 14:44

Mark Viola · Accepted Answer · 2015-11-14T16:24:57.437

10

HINT:

For $0<q<1$, let $q=1/(1+x)$ for $x>0$. Then, write $nq^n=\frac{n}{(1+x)^n}$ and apply the binomial theorem. This should facilitate a $\delta-\epsilon$ proof.

Can you proceed now?

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Using the binomial theorem we can write for $x>0$ \begin{align}(1+x)^n&=\sum_{k=0}^n\binom{n}{k}x^k\\\\&\ge \binom{n}{2}x^2\\\\&=\frac{n(n-1)}{2}x^2\end{align}Then, we have \begin{align}nq^n\le \frac{n}{\frac{n(n-1)}{2}x^2}=\frac{2}{(n-1)x^2}< \epsilon\end{align}whenever $n>1+\frac{2}{x^2\epsilon}$

edited Nov 14 '15 at 16:24

answered Nov 14 '15 at 14:33

Mark Viola

179,405

umm not really... – Yonatan Izutskiver Nov 14 '15 at 15:02
Do you know the binomial theorem? – Mark Viola Nov 14 '15 at 15:04
ive studied the binomial theorem, and now everything is crystal clear! besides one thing - what happens for $-1<q<0$? – Yonatan Izutskiver Nov 16 '15 at 19:28
If $-1<q<0$, then $0<-q<1$ and $\left|q^n\right|=\left|(-q)^n\right|$. And you can proceed now. – Mark Viola Nov 16 '15 at 19:45

Olivier Oloa · Answer 2 · 2015-11-14T14:23:58.710

1

The answer below was given before the OP changed his initial question.

It is sufficient to consider the case $0<q<1$. Then, if you know that, as $n \to +\infty$, $$ \frac{\ln n}{n} \to 0 $$ then you may just consider $$ \ln (nq^n)=\ln n+n\times \ln q=n \left(\underbrace{\ln q}_{\color{red}{<0}}+\underbrace{\frac{\ln n}{n}}_{\color{red}{\to 0}} \right) \to -\infty $$ implying that $$ nq^n \to 0, \quad n \to \infty. $$

edited Nov 14 '15 at 14:23

answered Nov 14 '15 at 14:03

Olivier Oloa

120,989

1

thanks for the answer, but i need to prove this statement, not to calculate the limit, using epsilon etc'... – Yonatan Izutskiver Nov 14 '15 at 14:09

Vlad Z · Answer 3 · 2015-11-15T10:24:06.033

ok, let's use the brute force approach here. To simplify my notation, I am going to assume non-negative $q$, but you can extend it easily to negative $q$ by substituting $q$ with $|q|$ below:

let's find $N$ that satisfies $Nq^n < \epsilon$

$q^N < \epsilon /N < \epsilon$

$N =\ln \epsilon / \ln q$

let's find $M$ after which $nq^n$ decreasing, that is:

$(M+1)q^{(M+1)} < Mq^M$

$q < \frac{M}{M+1} = 1 - \frac{1}{M+1}$

$\frac{1}{M+1} < 1- q$

$M > \frac{1}{1-q} -1$ (obviously $q < 1$)

That is for any $ n>M$, $nq^n$ has is smaller than its predecessor.

Lastly let's choose $N^* = max(N, M)$. Every $n > N^*$ satisfies the $\epsilon $ inequality, so are all further values of $n$, thus you have proven your case.

Prove that $\lim_{n\to∞}(nq^n)=0$, while $\lvert q\rvert <1$

3 Answers3