show that $\sum_{k=1}^{n}(1-a_{k})<\frac{2}{3}$ Let $a_{1}=\dfrac{1}{2}$, and such $a_{n+1}=a_{n}-a_{n}\ln{a_{n}}$,show that
$$\sum_{k=1}^{n}(1-a_{k})<\dfrac{2}{3}$$
My attemp: let $1-a_{n}=b_{n}$,then we have
$$b_{n+1}=b_{n}+(1-b_{n})\ln{(1-b_{n})}<b^2_{n}<\cdots<(b_{1})^{2^{n}}=\dfrac{1}{2^{2^n}}$$
where use $\ln{(1+x)}<x,x>-1$
so
$$\sum_{k=1}^{n}(1-a_{k})<\sum_{k=1}^{n}\dfrac{1}{2^{2^{k-1}}}?$$
But $$\sum_{k=1}^{+\infty}\dfrac{1}{2^{2^{k-1}}}=0.816\cdots$$big than$\frac{2}{3}$,so this inequality How to prove it?
 A: As used by the OP: $\enspace b_n:=1-a_n$
Examples: $\enspace\displaystyle b_1=0.5\enspace , \enspace b_2=0.5-0.5\ln 2\enspace , \enspace b_3 = b_2 + (1-b_2) \ln (1-b_2) <0.013$
We have: $\enspace b_n<-\ln(1-b_n)\enspace$ => $\enspace a_{n+1}=a_n(1-\ln a_n)>a_n(1+b_n)\enspace$ => $\enspace b_{n+1}<b_n^2\enspace$ 
Using $\,b_{n+1}<b_n^2\,$ we begin with $\,b_4=1-a_4<0.0001=0.1^{2^2}\,$ and get 
$\enspace b_n<0.1^{2^{n-2}}\,$ .
It follows:
$\displaystyle \sum\limits_{k=1}^n (1-a_k) < b_1 + b_2 + b_3 + \sum\limits_{k=4}^\infty 0.1^{2^{k-2}}$ 
$\displaystyle < 1- 0.5 \ln 2 + 0.013 + \sum\limits_{k=4}^5 0.1^{2^{k-2}} + \sum\limits_{k=4}^\infty 0.1^{12+k} $ 
$\displaystyle = 1- 0.5 \ln 2 + 0.013 + 0.00010001 + \frac{1}{9}0.1^{15} $
$\displaystyle < 1- 0.5 \ln 2 + 0.013 + 0.00010002 < 0,66652643 <\frac{2}{3}$
A: We can prove that $0< a_{n}< 1$ inductively by making use of the graph$:\quad y= x\left ( 1- \ln x \right ).$ We let
$$b_{n}:= 1- a_{n}, \left \{ b_{n} \right \}_{n= 1}^{\infty}\Leftrightarrow b_{1}= \frac{1}{2}, b_{n+ 1}= b_{n}+ \left ( 1- b_{n} \right )\ln\left ( 1- b_{n} \right )$$
Well$,\quad a_{n+ 1}- a_{n}= -a_{n}\ln n> 0,$ then $b_{n+ 1}< b_{n}\Rightarrow 0< b_{n}< b_{1}= \frac{1}{2},$ according to inequality
$$\ln\left ( 1- x \right )< -x,\quad x\in\left ( 0, 1 \right )$$
So$,\quad n> 2$
$$\Rightarrow b_{n}= b_{n- 1}+ \left ( 1- b_{n- 1} \right )\ln\left ( 1- b_{n- 1} \right )< b_{n- 1}- \left ( 1- b_{n- 1} \right )b_{n- 1}= b_{n- 1}^{2}\Rightarrow b_{n}< b_{2}^{2^{n- 2}}=$$
$$= \left ( \frac{1- \ln 2}{2} \right )^{2^{n- 2}}\Rightarrow\sum_{k= 1}^{n}b_{k}< \sum_{k= 1}^{n}\left ( \frac{1- \ln 2}{2} \right )^{2^{k- 2}}< \sum_{k= 1}^{\infty}\left ( \frac{1- \ln 2}{2} \right )^{2^{k- 2}}= 0.56\cdots< \frac{2}{3}$$
A: You can add more coefficients in the Taylor expansion
$$\log(1-x)\leq-x-x^2/2+...$$ 
in order to obtain the sharper bound
$$b_{n}+(1-b_{n})\ln{(1-b_{n})}<b^2_{n}/2+b_n^3/2$$
and so on, if it is necessary.
