Show using mathematical induciton that `n!lt ((n+1)/(2))^n`. Where `n in N and n gt 1`.

Explanation

Let `P(n):n! lt ((n+1)/(2))^n`
Step I For `n=2,2!lt ((2+1)/(2))^1rArr 2 lt (9)/(4)`
`rArr 2lt 2.25`, which is true.
Therefore , P(2)is true .
Step III For `n=k+1, we shall prove that `P(k+1):(k+1)!lt ((k+2)/(2))^(k+1)`
From assumption step `k!((k+1)^k)/(2^k)`
`rArr (k+1)k!lt ((k+1)^(k+1))/(2^k)`
`rArr (k+1)!lt ((k+1)^(k+1))/(2^k)` ......(i)
Let us assume , `((k+1)^(k+1))/(2^k)lt ((k+2)/(2))^(k+1)` .....(ii)
`rArr ((k+2)/(k+1))^(k+1)gt 2rArr (1+(1)/(k+1))^(k+1)gt 2`
`rArr 1+(k+1).(1)/((k+1))+.^(k+1)C_(2)((1)/(k+1))^2+....gt 2`
`rArr 1+1+.^(k+1)C_(2)((1)/(k+1))^2+.....gt 2`
Which is true , hence Eq. (ii) is true. From Eqs. (i) and (ii) , we get `(k+1)!lt ((k+1)^(k+1))/(2^k)` lt ((k+2)/(2))^(k+1)`
`rArr (k+1)!lt ((k+2)/(2))^(k+1)`
Therefore , `P(k+1)` is true. Hence , by the principle of mathematical induction P(n) is true for all `n in N`.