Proof of the Infinite Series That Calculates 'e'Date: 02/04/2004 at 15:56:36 From: Tarik Selbes Subject: proof of the infinite series which gives the constant e Is there a proof about this infinite series that gives the value of e: 1 + 1/1! + 1/2! + 1/3! + 1/4! + . . . + 1/n! where n goes to infinity? Date: 02/04/2004 at 21:18:13 From: Doctor Rob Subject: Re: proof of the infinite series which gives the constant e Thanks for writing to Ask Dr. Math, Tarik! There are two proofs that I know of. 1. The function e^x has derivative equal to itself. Then the Maclaurin series for any function which can be differentiated as many times as you like is f(x) = f(0)/0! + f'(0)*x/1! + f"(0)*x^2/2! + f"'(0)*x^3/3! + ... For f(x) = e^x, you have e^x = f(x) = f'(x) = f"(x) = f"'(x) = ... 1 = f(0) = f'(0) = f"(0) = f"'(0) = ... and the Maclaurin series for e^x is then e^x = 1/0! + x/1! + x^2/2! + x^3/3! + x^4/4! + ... Now set x = 1, and you get the series about which you asked. 2. The definition of e is e = lim (1+1/n)^n n->oo Consider the binomial expansion for n = 1, 2, 3, 4, 5, ... n (1+1/n)^n = SUM C(n,i)*(1/n)^i i=0 For i = 0, 1, 2, 3, ... one has C(n,i)*(1/n)^i = n!/[(n-i)!*i!*n^i] = (1)*(1-1/n)*(1-2/n)*...*(1-[i-1]/n)/i! whose limit as n grows without bound is 1/i!. Then n lim (1+1/n)^n = lim SUM C(n,i)*(1/n)^i n->oo n->oo i=0 oo = SUM lim C(n,i)*(1/n)^i i=0 n->oo oo e = SUM 1/i! i=0 Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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