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MT 2016 ANALYSIS I A Number Called e These supplementary notes by H A Priestley provide a (non-examinable) proof of the useful fact that n 1 e= 1+ . n [An alternative, and simpler, proof of the more general result in which x ∈ R>0 replaces n can be based on L’Hôpital’s Rule (in Analysis II).] e.1. The number e is defined to be ∞ X 1 1 1 e= = 1 + + + ··· , n! 1! 2! k=0 where, by convention, 0! = 1. Problem sheet 5, Q. 5, asked for a proof that the partial sum sequence of the series above is monotonic increasing and bounded above. Hence it converges to a real number, so that e is well defined. You were also asked to show e is irrational. Problem sheet 1, Q.5, introduced sequences n n+1 1 1 αn = 1 + and βn = 1 + n n and asked for a proof that, for all n, αn 6 αn+1 6 · · · 6 βn+1 6 βn . Example 6.3(c) then applied the Monotonic Sequences Theorem to prove that (αn ) converges. We now provide the desired reconciliation. e.2 Proposition. n 1 e = lim 1 + . n→∞ n n X 1 Proof. Let sn = . n! k=0 By the Binomial Theorem 2 3 n (n − 1) 1 n (n − 1) (n − 2) 1 1 1 αn = 1 + n + + + ··· + n n 2! n 3! n n 1 1 1 1 2 1 1 2 1 =1+1+ 1− + 1− 1− + ··· + 1− 1− ··· 2! n 3! n n n! n n n 1 1 1 6 1 + 1 + + + ··· + = sn . 2! 3! n! From this we have lim αn 6 e. 1 2 On the other hand, if m, n are natural numbers with m < n, focusing on the first m + 1 terms of αn we see that 1 1 1 2 m−1 1 1+1+ 1− + ··· + 1 − 1− ··· 1 − 6 αn . n 2! n n n m! If we fix m and let n → ∞ then we have, using the Algebra of Limits and recalling that limits respect weak inequalities, 1 1 1 1 + 1 + + + ··· + 6 lim αn . 2! 3! m! Finally letting m → ∞ we have e 6 lim αn and the result follows. e.3 Another useful limit. n 1 1 1− → . n e Proof. See Problem sheet 5, Q. 6.