I have the following inequality $$5*10^{-10} \geq \dfrac{2^{n+1}}{(n+1)!}$$
Is there any way this can be solved algebraically? If not, is there a method that is better than guessing, for finding the smallest $n$ which satisfies the inequality?
I have the following inequality $$5*10^{-10} \geq \dfrac{2^{n+1}}{(n+1)!}$$
Is there any way this can be solved algebraically? If not, is there a method that is better than guessing, for finding the smallest $n$ which satisfies the inequality?
On
It might help to think of this as a ratio to get a (very) rough estimate:
When does the factorial have 10 times as many digits as 2^n?
Factorial grows a bit faster than one digit per n (for small n), 2^n grows a bit faster than three digits per 10 n (2^10 = 1024). Then:
$ n = \frac{3}{10}n + 10 $
estimate is n=14 (not n+1, as in the original equation).
Or you could just count the number of digits, which shows n ~ 18
n factorial digits 2^n digits difference
-----------------------------------------------------------------
0 1 1 1 1 0
1 1 1 2 1 0
2 2 1 4 1 0
3 6 1 8 1 0
4 24 2 16 2 0
5 120 3 32 2 1
6 720 3 64 2 1
7 5040 4 128 3 1
8 40320 5 256 3 2
9 362880 6 512 3 3
10 3628800 7 1024 4 3
11 39916800 8 2048 4 4
12 479001600 9 4096 4 5
13 6227020800 10 8192 4 6
14 87178291200 11 16384 5 6
15 1307674368000 13 32768 5 8
16 20922789888000 14 65536 5 9
17 355687428096000 15 131072 6 9
18 6402373705728000 16 262144 6 10
19 121645100408832000 18 524288 6 12
20 2432902008176640000 19 1048576 7 12
this is a method from which you can calculate the value of $n$ , but it need not be the most accurate , $\frac{2^{n+1}}{(n+1)!}$<2*($\frac{2}{3})^n$<$5*10^{-10}$ now using log you can calculate the value of $n$ , but note that is not the smallest $n$