Re: Analysis with continuous, sum.

In article <gdi4br$c8v$1@xxxxxxxxxxxxxxxxxxxxx>,
"mina_world" <mina_world@xxxxxxxxxxx> wrote:

Hello teacher~

f(x) = sum{n=1 to oo} (x^n)(1-x) / n

Show that f(x) is continuous on [0, 1].

The maximum value of (x^n)(1-x)/n on [0,1] is (n/(n+1))^n *
[1-(n/(n+1)]/n <= 1/n^2. Because sum 1/n^2 < oo, your series converges
uniformly on [0,1] by the Weierstrass M test. The partial sums are all
continuous on [0,1], hence so is f.

--------------------------------------------------
Hm...maybe,
I must show that f(x) converges uniformly on [0, 1] ?

Hm...
For n >= m > N,
|f_n(x) - f_m(x)| =
|{(x^(m+1))(1-x) / (m+1)} + ... + {(x^n)(1-x) / n}|
<= (1-x).|{x^(m+1) / (m+1)} + ... + {(x^n) / (m+1)}|
<= (n-m)(x^(m+1)) / (m+1)
<= (n-m) / (m+1)
??

Maybe, I need your rebuke.
.