Proving that $n \over 2^n$ converges to $0$ I'm completely clueless on this one.
I can easily calculate the limit using L'Hopital's rule, but proving that the series is converging to 0 is far more tricky.
$$a_n= {n \over 2^n}$$ 
Any help?
 A: Using the fact that $2^n>n^2$ for $n > 4$, we have: $$0 \leq a_n < \frac{n}{n^2}=\frac{1}{n}.$$
Hence, $\displaystyle \lim_{n \to \infty}a_n=0.$
A: By the binomial theorem, for $n\ge 2$,
$$2^n=(1+1)^n\ge \binom n0+\binom n1+\binom n2=1+n+\frac{n(n-1)}{2}$$
$$\Rightarrow \frac{n}{2^n}\le \frac{n}{1+n+\frac{n(n-1)}{2}}$$
A: The series
$$\sum_{n=1}^\infty n x^n$$
converges for $|x| < 1$. Hence we must have
$$ \lim_{n \to \infty} \left( n x^n \right) = 0$$
Choose $x = \frac{1}{2} < 1$ and you result follows!
A: A really elementary way is to observe the following:
$$a_n=\frac{n}{2^n}=\frac{n}{n-1}\frac{2^{n-1}}{2^n}\frac{n-1}{2^{n-1}}=\frac{n}{n-1}\frac{2^{n-1}}{2^n}a_{n-1}=\frac{n}{2(n-1)}a_{n-1}.$$
Let $c_n:=\frac{n}{2(n-1)}$. Prove for yourself that $c_n$ is smaller than $3/4$ whenever $n\geq 3$. Then what we have shown is
$$a_n\leq \frac{3}{4}a_{n-1}\leq\dots\leq (\frac{3}{4})^na_3.$$
Conclude!
A: We see that, $$a_{n+1}-a_n=\frac{n+1}{2^{n+1}}-\frac{n}{2^n}$$
$$=\frac{n+1-2n}{2^{n+1}}=\frac{1-n}{2^{n+1}} \le 0$$
This implies that {$a_n$} is strictly monotonic decreasing.
Again $a_n=\frac{n}{2^n}>0 \,\ \forall \,\ n \in \mathbb{N}$
Hence {$a_n$} is bounded below by $0$.
So, {$a_n$} is convergent and converges to greatest lower bound, in this case, $0$. You can show this limit in several ways.
