limit question: $\lim\limits_{n\to \infty } \frac{n}{2^n}=0$

Let's do something different!!

Note that the sequence $\{\frac{n}{2^n}\}_{n\geq 1}$ is decreasing ( easy to prove) and bounded from below by $0$, thus $\lim_{n\rightarrow \infty}\frac{n}{2^n}$ exists. Call it $L$.

$$L=\lim_{n\rightarrow \infty}\frac{n}{2^n}=\lim_{n\rightarrow \infty}\frac{(2n)}{2^{(2n)}}=\lim_{n\rightarrow\infty}(\frac{1}{2^{n-1}}\frac{n}{2^n})=[\lim_{n\rightarrow\infty}\frac{1}{2^{n-1}}][\lim_{n\rightarrow \infty}\frac{n}{2^n}]=(0)(L)=0$$

Put

$$a_n:=\frac{\sqrt[n]n}2\implies a_n\xrightarrow [n\to\infty]{}\frac12\implies$$

If we take say $\,\epsilon=0.1\;$ then

$$\exists\,N_\epsilon\in\Bbb N\;\;s.t.\;\;n>N_\epsilon\implies \left|a_n-\frac12\right|<0.1\iff \frac25<a_n<\frac35\implies$$

$$\implies \left(\frac25\right)^n<a_n^n=\frac n{2^n}<\left(\frac35\right)^n$$

Now apply the squeeze theorem and get what you want

Let's be original. Let $$ \frac 1{1-x} = \sum_{k \ge 0} x^k, $$ which means the derivative of this series is also convergent with the same radius of convergence (namely, $1$) by Taylor's theorem : $$ x \left( \frac 1{(1-x)^2} \right) = x \left( \sum_{k \ge 0} k x^{k-1} \right) = \sum_{k \ge 0} k x^k. $$ Letting $x = 1/2$, we see that the series $$ \sum_{k \ge 0} \frac{k}{2^k} $$ is convergent, hence $\lim_{k \to \infty} \frac{k}{2^k} = 0$.

Hope that helps,