Re: Limit of a sequence

In article
<4417266.1208321911927.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
TCL <tlim1@xxxxxxx> wrote:

In article
<15986032.1208228781859.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
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.

Factor out exp(-k/n) and then ignore it: We're left with

Sum_{k=1 to n} (1/k)*(1 - exp(k/n)*(1-1/n)^k) (1).

Using ln(1-x) = - x + O(x^2) for small x, we have

ln(exp(k/n)*(1-1/n)^k) = k/n + k(-1/n + O(1/n^2)) = k*O(1/n^2).

Exponentiating back, using e^x = 1 + O(x) for x bounded gives

exp(k/n)*(1-1/n)^k = 1 + k*O(1/n^2).

So (1) = Sum_{k=1 to n} (1/k)*k*O(1/n^2) = O(1/n).
.

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