How do we show $\lim_{n\to \infty}\frac{n+1}{2^{(n+1)}}=0$? How do we show $\lim_{n\to \infty}\frac{n+1}{2^{(n+1)}}=0$?
It is clear that the denominator increases faster than the numerator, but is there a rigorous way of stating it. i.e. using a limit law.
I tried to cancel by $n$ and get $\frac{1+\frac1n}{\frac{2^{(n+1)}}n}$, but that doesn't quite help.
 A: If $a_n=\frac{n}{2^n}$, then $a_1=\frac{1}{2}=a_2$ and
$$ \frac{a_{n+1}}{a_n}=\frac{n+1}{2^{n+1}}\frac{2^n}{n}=\frac{1}{2}\cdot\frac{n+1}{n}\leq \frac{3}{4}$$
for $n\geq 2$. Therefore
$$ 0\leq a_n\leq \frac{3}{4}a_{n-1}\leq\cdots\leq \Big(\frac{3}{4}\Big)^{n-2}a_2=\frac{1}{2}\Big(\frac{3}{4}\Big)^{n-2}$$
and so it follows that $a_n\to 0$ by the squeeze theorem.
A: You could treat $n$ as a real variable and use L'Hospital rule 
A: You can prove that $2^n>n^{2}$ for all $n\geq 5$ (perhaps via induction) and then we can see that $0<n/2^{n}<n/n^{2}=1/n$ and by squeeze theorem the desired limit is $0$.
In general if $|x|<1$ then $n^{r} x^{n} \to 0$ no matter how large the exponent $r$ is. This is proved either using the ratio test (as done in the answer by user carmichael561) or by using another fundamental limit $(\log n) /n\to 0$.
A: Note that
$$ 2^{n+1}=1+\binom{n+1}{1}+\binom{n+1}{2}+\cdots+\binom{n+1}{n+1}\ge\binom{n+1}{2}=\frac12(n+1)(n+2)$$
and hence
$$ 0<\frac{n+1}{2^{n+1}}\le\frac{2}{n+2}.$$
It is easy to see
$$\lim_{n\to\infty}\frac{n+1}{2^{n+1}}=0.$$
