Re: Limit of a sequence

In article <20080415.115000@xxxxxxxx>,
rob@xxxxxxxxxxxxxx (Rob Johnson) wrote:

In article <aderamey.addw-6681CD.10293515042008@xxxxxxxxxxxxxxxxxxxxxx>,
The World Wide Wade <aderamey.addw@xxxxxxxxxxx> wrote:
In article <20080415.014147@xxxxxxxx>,
rob@xxxxxxxxxxxxxx (Rob Johnson) 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.

Not quite a Riemann sum for that. Maybe a "Riemann-type" sum.

Yes, what I should have said is that as n -> oo, (1 - 1/n)^n -> 1/e,
so the summation above and the summation below tend to the same limit.

n
--- n -k/n 1
> - ( 1 - e ) -
--- k n
k=1

This last summation is a Riemann sum for the integral above.

This integral can be reduced using power series to

oo
--- (-1)^{k-1}
> ----------
--- k k!
k=1

which evaluates to approximately .796599599297.

Another way to see that integral is < 1 is to note that for each x in
(0,1], (1 - e^(-x))/x = e^(c_x) for some c_x in (0,x) by the mean
value theorem. Hence the integrand < 1 on (0,1].

I think you meant either "e^(-c_x)" or "for some c_x in (-x,0)", but
yes, that works too.

Right, I meant the first.
.

Relevant Pages

  • Re: Limit of a sequence
    ... Not quite a Riemann sum for that. ... so the summation above and the summation below tend to the same limit. ... Hence the integrand < 1 on (0,1]. ... Rob Johnson ...
    (sci.math)
  • Re: Limit of a sequence
    ... TCL wrote: ... this is not quite a Riemann sum. ... To build an epsilon-delta proof we will ... Rob Johnson ...
    (sci.math)
  • Re: Limit of a sequence
    ... TCL wrote: ... with x = k/n. ... Not quite a Riemann sum for that. ... Hence the integrand < 1 on (0,1]. ...
    (sci.math)