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Monty Phister guest
May-29-08, 12:31 PM (EST)

"Infinite series for e"

 I think I know that e=lim(n->infinity) (1+1/n)^nAnd the ith term of this series is 1/(i-1)!But the ith term is also n(n-1)(n-2)(n-3)...(n-(i-2))/(n^(i-1))(i-1)!It would seem, then, that as n->infinity, the term n(n-1)(n-2)(n-3)...(n-(i-2))/n^(i-1) must approach 1, so we're left with 1/(i-1)!.I can't seem to prove that this is true, and in fact taking various examples with i roughly half of n and n growing, the term gets smaller as n gets larger.Why am I confused? No, correct that. I'm usally confused. What I mean is, where's the error in my logic or calculation?

alexb
Charter Member
2229 posts
May-29-08, 12:56 PM (EST)    1. "RE: Infinite series for e"
In response to message #0

 >I think I know that e=lim(n->infinity) (1+1/n)^n That is correct.>And the ith term of this series is 1/(i-1)! You mean the ith term of e^1, right?>But the ith term is also > n(n-1)(n-2)(n-3)...(n-(i-2))/(n^(i-1))(i-1)! "Also" is inappropriate here. At best you may claim that the limit of n(n-1)(n-2)(n-3)...(n-(i-2))/(n^(i-1))(i-1)! as n goes to infinity is 1/(i-1)!.I would accept that, yes.>It would seem, then, that as n->infinity, the term > n(n-1)(n-2)(n-3)...(n-(i-2))/n^(i-1) must approach 1, Yes, I would expect that.>I can't seem to prove that this is true, lim (n - a)/n = 1for any a. The limit of a finite product of such ratios is also 1:lim (n - a)/n · (n - b)/n · ... · (n - c)/n = 1,for given (and independent of n) a, b, ..., c.>and in fact taking various examples with >i roughly half of n and n growing, the >term gets smaller as n gets larger. You can't take i "roughly half of n". i must be fixed as n grows.>Why am I confused? No, correct that. I'm usally confused. >What I mean is, where's the error in my logic or >calculation? If i is "roughly half of n" then obviously the ith term must tend to 0. For, otherwise, the sum of an infinite number of such terms would not be finite. Monty Phister guest
May-30-08, 11:30 AM (EST)

2. "RE: Infinite series for e"
In response to message #1

 Thanks, Alex. As you say, the 'also' is inappropriate. I think my difficulty is that I was somehow thinking there was a last term to the series expansion with n-> infinity. And your very last comment clears up the misconception in my calculations.

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