Re: Limit of a sequence
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Wed, 16 Apr 2008 12:45:02 GMT
In article <4417266.1208321911927.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
TCL <tlim1@xxxxxxx> wrote:
In article <15986032.1208228781859.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
TCL <tlim1@xxxxxxx> wrote:
For positive integers n, define
s_n = sum_{k=1 to n} of (1/k)* (1-(1-1/n)^k).
Prove that s_n --> 1 as n-->infty.
n
--- n k 1
> - ( 1 - (1 - 1/n) ) -
--- k n
k=1
is a Riemann Sum for
|\1 1 -x
| - ( 1 - e ) dx
\| 0 x
with x = k/n.
As noted by WWW, this is not quite a Riemann sum. To justify your steps, one needs to prove
Sum_{k=1 to n} (1/k)* ( exp(-k/n) - (1-1/n)^k )
converge to 0 as n-->infty. The fact that lim (1-1/n)
^n=e^{-1} may be helpful, but that does not prove that
the above limit is 0.
Indeed it does prove it. To build an epsilon-delta proof we will
use the following simple estimate.
By the mean value theorem, we have for some c in [a,b]
a^x - b^x x-1
--------- = x c [1]
a - b
If x is in [0,1] and both a and b are between (1 - 1/2)^2 and 1/e,
then we have that c^{x-1} < 4. Under these conditions, we have
|a^x - b^x| < 4 x |a - b| [2]
Now, assuming only that (1 - 1/n)^n -> 1/e, we can proceed.
For any d > 0, choose N so that if n > N, |(1 - 1/n)^n - 1/e| < d/4.
For n > N, we have that
n
--- n k/n n k/n 1
> - | (1/e) - ((1 - 1/n) ) | -
--- k n
k=1
n
--- n k 1
<= > - (4 - d/4) -
--- k n n
k=1
= d
Rob Johnson <rob@xxxxxxxxxxxxxx>
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