Proof that $\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$ Please show how to prove that $$\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$$
 A: Notice that
$$
\frac{(n+1)/2^{n+1}}{n/2^n}=\frac12\left(1+\frac1n\right)\tag{1}
$$
For $n\ge2$, the ratio in $(1)$ is at most $\frac34$. At $n=2$, $\dfrac{n}{2^n}=\dfrac12$. Therefore, $(1)$ implies
$$
\frac{n}{2^n}\le\frac12\left(\frac34\right)^{n-2}\tag{2}
$$
for $n\ge2$. Hopefully, it is clearer that
$$
\lim_{n\to\infty}\frac12\left(\frac34\right)^{n-2}=0\tag{3}
$$
A: By the Binomial Theorem, if $n\ge 2$ then
$$(1+1)^n \ge 1+n+\frac{n(n-1)}{2}\gt \frac{n(n-1)}{2}.$$
It follows that if $n\ge 2$ then $\dfrac{n}{2^n}\lt \dfrac{2}{n-1}$. 
A: With Stolz-Cesaro Theorem :
$$
\color{#66f}{\large\lim_{n \to \infty}{n \over 2^{n}}}
=\lim_{n \to \infty}{\left(n + 1\right) - n \over 2^{n + 1} - 2^{n}}
=\lim_{n \to \infty}{1 \over 2^{n}}=\color{#66f}{\Large 0}
$$
A: Consider extending the sequence {$n/2^n$} to the function $f(x)=x/2^x$.  
Then use L'Hopital's rule: lim$_{x\to\infty} x/2^x$ has indeterminate form $\infty/\infty$. Taking the limit of the quotient of derivatives we get lim$_{x\to\infty} 1/($ln$2\cdot 2^x)=0$.  Thus lim$_{x\to\infty} x/2^x=0$ and so $n/2^n\to 0$ as $n\to\infty$.
A: Elementary (?) Proof.    
$n\geq4$ $~\Longleftrightarrow~$ $\dfrac{n}{n-1}<\sqrt{2}$    
$\dfrac{n}{4}=\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\cdots\times\dfrac{n}{n-1}<\sqrt{2}\times\sqrt{2}\times\sqrt{2}\cdots\times\sqrt{2}=2^{\left(\frac{n}{2}-2\right)}$    
$\therefore$ $n^{2}<2^{n}$$~\Longrightarrow~$ $0<\dfrac{n}{2^{n}}<\dfrac{1}{n}$    
$\displaystyle 0=\lim_{n\to\infty}0\leq\lim_{n\to\infty}\dfrac{n}{2^{n}}\leq\lim_{n\to\infty}\dfrac{1}{n}=0$    
$\therefore$ $\displaystyle\lim_{n\to\infty}\dfrac{n}{2^{n}}=0$
