Is there some sort of approximation for $2^n$? I'm specifically interested in how the ratio $\frac{2n}{2^n}$ scales for large $n$ (apart from decreasing to zero in the limit)
Asked
Active
Viewed 165 times
0
-
Maybe this will help? http://mathcentral.uregina.ca/QQ/database/QQ.09.06/h/nick1.html Conclusion: Logarithms. – Matti P. Aug 05 '19 at 11:42
-
You have $\frac{n}{2^{n-1}}$ which gets very small as n goes large. In fact as n approaches infinity, the expression approaches zero. – NoChance Aug 05 '19 at 11:50
-
1I suggest you clarify your question. – NoChance Aug 05 '19 at 12:02
-
3The usual approximation for $2^n$ is $2^n$. – Gerry Myerson Aug 05 '19 at 12:28
1 Answers
1
Let $r_n = \frac{2n}{2^n}$. Then $$ \frac{r_{n+1}}{r_n}=\frac12\left(1+\frac1n\right) \to \frac12 $$ In words, for large $n$, we have that $r_{n+1}$ is approximately half of $r_n$ but slightly greater.
lhf
- 216,483