Prove $\displaystyle\lim_{n\to \infty}{nb^n}=0, 0<b<1$

One last thing I used a certain theorem in my proof which I will give.

Theorem 3.1.10 Let $(x_n)$ be a sequence of real numbers and let $x\in \mathbb{R}$ If $(a_n)$ is a sequence of positive real numbers with $\lim(a_n) = 0$ and if for some constant $C > 0$ and some $m \in\mathbb{N}$ we have $|x_n-x|\leq Ca_n$ for all $n > m$, then it follows that $lim(x_n ) = x$.

Here is how my proof goes. Though I don't think its right.

Since $0<b<1$ then we can write $b=\frac{1}{1+h_n}$ for some $h_n>0$ and $n \in \mathbb{N}$ It follows that $nb^n=\frac{n}{(1+h_n)^n}$. Then by Bernoulli's inequality we get $\frac{n}{(1+h_n)^n}\leq \frac{n}{(1+nh_n)}\leq \frac{n}{nh_n}$ By the theorem above we let $C=\frac{n}{h_n}$ and $a_n=\frac{1}{n}$. Thus $\displaystyle\lim_{n\to \infty}{nb^n}=0$

  • $\begingroup$ I think im gonna do it another way around. $\endgroup$ – user60887 Jun 29 '13 at 0:10
  • $\begingroup$ I was slightly mistaken. I don't know why you wrote $b=1+h_n$, since $b$ is fixed. The problem is that with the $n$ factor above, you need to take a little more in the binomial expansion. $\endgroup$ – Pedro Tamaroff Jun 29 '13 at 0:12
  • $\begingroup$ so is my proof almost correct? $\endgroup$ – user60887 Jun 29 '13 at 0:13
  • $\begingroup$ Well, not really. $C=\dfrac nh$ is not a constant. $\endgroup$ – Pedro Tamaroff Jun 29 '13 at 0:14
  • $\begingroup$ oh ok im going to fix it then. $\endgroup$ – user60887 Jun 29 '13 at 0:15

Equivalently, we want to prove the following: $$\lim_{n\to\infty}\frac{n}{\alpha^n}=0$$ whenever $\alpha >1$. I will use your idea. Since $\alpha >1$; we can write $\alpha=1+h$ for $h>0$. Then $$(1+h)^n\geq 1+hn+\frac{h^2}{2}n(n-1)$$

eventually (for $n\geq 2$)

Thus, we can write $$\frac{n}{\alpha^n}\leq \frac{n}{P_2(n)}$$

where $P_2$ is a polynomial of degree $2$. In general, we can prove: $$\lim_{n\to\infty}\frac{n^k}{\alpha^n}=0$$

whenever $\alpha >1$ and $k$ is any positive fixed integer. We use the same as above, but take up to the $k+1$-th term. Then we get $$\frac{n^k}{\alpha^n}\leq \frac{n^k}{P_{k+1}(n)}$$ where $P_{k+1}$ is a polynomial of degree $k+1$.


For one thing, you don't need to write $h_n$ - $h$ will do since it doesn't vary.

I think your proof is incorrect.

You have, using the preceding comment, $\frac{n}{(1+h)^n}\leq \frac{n}{(1+nh)}\leq \frac{n}{nh} = \frac1{h} $, so you only show that the terms are bounded, not go to zero.

Here is an approach that seems to work using Bernoulli's inequality:

Write $n = 2m$, since we are concerned with large $n$. Then $(1+h)^n = (1+h)^{2m} (1+h)^m(1+h)^m \ge (1+hm)(1+hm) > h^2 m^2 = h^2 n^2/4 $, so $\dfrac{n}{(1+h)^n} < \dfrac{n}{h^2 n^2/4} = \dfrac{4}{h^2n} \to 0 $.

This generalizes to showing $\dfrac{n^k}{(1+h)^n} \to 0$ like this:

Let $n = m(k+1)$. Choose an integer $m > k$. (If you want to be able to use any $n$, not just multiples of $k$, you need Bernoulli's inequality for fractional exponents, not just integral exponents.)


$\begin{align} (1+h)^n &= (1+h)^{m(k+1)}\\ &=((1+h)^m)^{k+1}\\ &> (1+mh)^{k+1}\\ &>(mh)^{k+1}\\ &=h^{k+1} (n/(k+1))^{k+1}\\ &=(h/(k+1))^{k+1} n^{k+1}\\ \end{align} $

so $\dfrac{n^k}{(1+h)^n} < \dfrac{n^k}{(h/(k+1))^{k+1} n^{k+1}} =(\dfrac{k+1}{h})^{k+1}\dfrac1{n} \to 0 $ since $(\dfrac{k+1}{h})^{k+1}$ is constant for fixed $b$ and $k$.


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