Solution of Problem 1, Section 6
Mathematical Method in the Physical Sciences 3th - Mary L. Boas
Question:
Show that n! > 2^n for all n > 3. Hint: Write out a few terms; then consider what you multiply by to go from, say 5! to 6! and from 2^5 to 2^6.
Answer:
To show that n! > 2^n for all n > 3, we can proceed as follows:
1. Write out the first few terms of both n! and 2^n to get a sense of their behavior:
n = 4: 4! = 24, 2^4 = 16
n = 5: 5! = 120, 2^5 = 32
n = 6: 6! = 720, 2^6 = 64
2. Consider what happens when we multiply by n+1 to go from n! to (n+1)! and from 2^n to 2^(n+1):
For n!, we multiply by n+1 to get (n+1)!. This means that (n+1)! = n! * (n+1) > n!
For 2^n, we multiply by 2 to get 2^(n+1). This means that 2^(n+1) = 2^n * 2 = 2 * 2^n < 2^n
Therefore, n! > 2^n, and n! will continue to grow faster than 2^n as n increases.
Therefore, we can conclude that for all n > 3, n! > 2^n
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