Proof that $\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$ Please show how to prove that $$\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$$

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
$$
\lim_{n\to\infty}\frac12\left(\frac34\right)^{n-2}=0\tag{3}
$$
