Real Analysis, limit of function using binomial theorem. We have $b \in \mathbb{R}$ and $ 0<b<1$, I need to show that $\lim(nb^n)=0$ 
My attempt:
Since $0<b<1$ we can express b as $\frac{1}{1+a_n}$ so now our sequence is $nb^n=n\left(\frac{1}{1+a_n}\right)^n=\frac{n}{(1+a_n)^n}$, we can use the binomial theorem to express $b^n=\frac{1}{1+na_n+\frac{1}{2}n(n-1)a_n^2+...}\leq\frac{1}{1+na_n+\frac{1}{2}n(n-1)a_n^2}$ So we have $nb^n \leq \frac{n}{1+na_n+\frac{1}{2}n(n-1)a_n^2}$. 
I am stuck after this, I am uncertain how to keep going from there. 
Thanks! 
 A: You are almost there with your method. Let $b = \dfrac1{1+a}$, where $a>0$, as you have done. We then have
$$b^n = \dfrac1{(1+a)^n} = \dfrac1{\displaystyle \sum_{k=0}^n \dbinom{n}k a^k} < \dfrac1{\displaystyle \sum_{k=2}^2 \dbinom{n}k a^k} = \dfrac2{n(n-1)a^2}$$
Hence, we have
$$nb^n < \dfrac2{(n-1)a^2}$$
Now let $n \to \infty$ and conclude what you want.
A: Potential method 1: Show that $\sum_{n=1}^{\infty} n b^n < \infty$ using the ratio test, this will imply that $nb^n \to 0$. 
Potential method 2: Notice that $\log(nb^n) = \log(n)-\lambda n$ where $\lambda = |\log(b)|>0$. If you know that $n >> \log(n)$ then you have that $\log(nb^n) \to -\infty$ as $n \to \infty$ implying (by continuity of $e^x$) that $nb^n = e^{\log(nb^n)} \to 0$.
A: Since $0<b<1, \ b^x$ is uniformly continuous, you can interchange limit function and differentiation. Hence 
$$
\lim_{x\ to \infty} x b^x = \frac{1}{b}\lim_{x \to \infty} x b^{x-1}=\frac{1}{b} \lim_{x \to \infty}\frac{\partial b^x}{\partial b}=\frac{1}{b}\frac{\partial}{\partial b}\lim_{x \to \infty}b^x=0
$$ 
Sorry no Binomial theorem here
