Something around the number e 1 n 1 Show that the sequence 1 n converges and denote the limit by e Proof Since n 1 1 n n nk k 0 1 n k n n 1 1 2 n n 1 1 1 n 1 n 1 n n n 2 n 1 1 n 1 1 1 1 1 1 n n 1 n 1 n 2 1 1 1 1 12 n 1 2 2 2 3 and by 1 we know that the sequence is increasing Hence the sequence is convergent We denote its limit e That is n lim 1 1 e n n 1 Remark 1 The sequence and e first appear in the mail that Euler wrote to Goldbach It is a beautiful formula involving e i 1 0 2 Use the exercise we can show that k 0 1 k e as follows n Proof Let x n 1 1n and let k n we have 1 1 1 1 1 1 1 1 1 n 1 n 2 k k k which implies that let k xk n i 1 y n e 2 i 0 On the other hand xn yn 3 So by 2 and 3 we finally have k 1 e 4 k 0 3 e is an irrational number Proof Assume that e is a rational number say e p q where g c d p q 1 Note that q 1 Consider q e q k 1 k 0 q q k 0 and since q integer However q 1 k 0 k 1 k q k q 1 1 k and q e are integers we have q k q 1 1 k is also an q k q 1 1 k k q 1 q k 1 1 q 1 q 1 q 2 2 1 1 q 1 q 1 1 q 1 a contradiction So we know that e is not a rational number n 4 Here is an estimate about e k 0 that e 2 71828 18284 59045 Proof Since e k 0 1 k n n where 0 1 In fact we know we have 0 e xn k n 1 1 k 1 where x n k n k 1 k 0 1 1 1 1 n 2 n 1 n 2 n 3 1 1 1 1 n 2 n 1 n 2 2 1 n 2 n 1 n 1 1 since n 2 2 1 n n n n 1 So we finally have n e k 1 k 0 where 0 1 n n Note We can use the estimate dorectly to show e is an irrational number 2 For continuous variables we have the samae result as follows That is x lim 1 1x e x n Proof 1 Since 1 1n e as n we know that for any sequence a n N with a n we have an lim 1 a1n e n 2 Given a sequence x n with x n and define a n x n then a n x n a n 1 then we have an xn a n 1 1 1 1 x1n 1 a1n an 1 Since an a n 1 1 1 e and 1 a1n e as x by 5 an 1 we know that 5 1 x n e lim 1 xn n Since x n is arbitrary chosen so that it goes infinity we finally obtain that 1 x e lim 1 x x 1 x 3 In order to show 1 x e as x we let x y then x 1 y 1 1x 1 y y y y 1 6 y 1 1 1 1 y 1 y 1 Note that x y by 6 we have shown that 1 1 y 1 x x lim 1 1x e y lim 1 3 Prove that as x 0 we have 1 dstrictly ecreasing 1 x y 1 1 1 y 1 x is strictly increasing and 1 1 x x 1 is Proof Since by Mean Value Theorem 1 log 1 1 log x 1 log x 1 1 for all x 0 x x x 1 we have x log 1 1x log 1 1x 1 0 for all x 0 x 1 and log 1 1x 1x 0 for all x 0 x 1 log 1 1x Hence we know that x log 1 1x is strictly increasing on 0 and x 1 log 1 1x is strictly decreasing on 0 It implies that x x 1 1 1x is strictly increasing 0 and 1 1x is strictly decreasing on 0 Remark By exercise 2 we know that 1 x e lim 1 1 1 lim x x x x x 1 4 Follow the Exercise 3 to find the smallest a such that 1 decreasing for all x 0 Proof Let f x 1 1 x x a 1 x and consider log f x x a log 1 1x Let us consider g x x a e and strictly x2 a x x log 1 y y 1 a y 2 g x log 1 1x 1 where 0 y 1 1 1 x 1 y yk y 1 a y 2 y k k k 0 k 1 1 a y 2 1 a y 3 1 a y n n 2 3 It is clear that for a 1 2 we have g x 0 for all x 0 Note that for a 1 2 if there exists such a so that f is strictly decreasing for all x 0 Then g x 0 for all x 0 However it is impossible since n g x 1 a y 2 1 a y 3 1 n a y 2 3 1 a 0 as y 1 2 So we have proved that the smallest value of a is 1 2 x 1 2 Remark There is another proof to show that 1 1x is strictly decreasing on 0 Proof Consider h t 1 t and two points 1 1 and 1 1x 1 1 1 lying on the graph x From three areas the idea is that The area of lower rectangle The area of the curve The area of trapezoid So we have 1 1 x 1 x 1 1 log 1 1x 1 x 1 1 1 2x 1 1 x x 1 2 1 x x 1 Consider 1 1x x 1 2 1 1x x 1 2 log 1 1x x 1 2 1 x x 1 0 by 7 x 1 2 hence we know that 1 1x is strictly decreasing on 0 x Note Use the method of remark we know that 1 1x is strictly increasing on 0 7