Prove that `2 le (1+ (1)/(n))^(n) lt 3` for all `n in N`.

Explanation

Let `a_(n) = l(1+(1)/(n))^(n)`
For `n = 1, (1+1/n)^(n) = 2`
Now, `(1+1/n)^(n) = .^(n)C_(0) + .^(n)C_(1) (1/n) + .^(n)C_(2)(1/n)^(n) + "……." + .^(n)C_(r)(1/n)^(r ) + "……" + .^(n)C_(n)(1/n)^(n)`
` = 1+1+(n(n-1))/(2!) (1)/(n^(2)) + (n(n-1)(n-2))/(3!) = 1/(n^(3)) + "......"`
` + (n(n-1)xx"......"xx2xx1)/(n!) (1)/(n^(n)) " " (1)`
` = 2+(1)/(2!)(1-(1)/(n)) + (1)/(3!)(1+(1)/(n))(1-(2)/(n)) +"....."`
`+ (1)/(n!)(1-(1)/(n))(1-(2)/(n))"......"(1-(n-1)/(n))" "(2)`
Hence, `a_(n) ge 2 ` for all `n in N`.
Also, `a_(n) le 1 +1 + (1)/(2!) + (1)/(3!) + "....." + (1)/(r!) + "......" + (1)/(n!)`
Fo` 2 le r le n`, we have `r! = 1 xx 2 xx 3 xx "......" xx r ge 2^(r-1)`.
`:. a_(n) le 1 + 1 + 1/2 + 1/(2^(2))+ "......" + (1)/(2^(r-1))+"....."+(1)/(2^(n-1))`.
` = 1+(1-(1//2)^(n))/(1-(1//2))`
` = 1+2(1-1/(2^(n))) = 3 - (1)/(2^(n-1))`
`:. a_(n) le 3 - 1/(2^(n-1)) lt 3 AA n ge 1`