Prove $\lim_{n\to \infty}{nb^n}=0$ 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 $\displaystyle\lim_{n\to \infty}{nb^n}=0$ 
 A: 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 $k+1$.
A: 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.)
Then
$\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 $k$.
