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Subject: "Infinite series for e"     Previous Topic | Next Topic
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Monty Phister
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May-29-08, 12:31 PM (EST)
 
"Infinite series for e"
 
   I think I know that e=lim(n->infinity) (1+1/n)^n
And 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?


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alexb
Charter Member
2229 posts
May-29-08, 12:56 PM (EST)
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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 = 1

for 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.


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Monty Phister
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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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