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
$$\sum_{k=1}^{n}(1-a_{k})&lt;\dfrac{2}{3}$$

My attemp: let $1-a_{n}=b_{n}$,then we have
$$b_{n+1}=b_{n}+(1-b_{n})\ln{(1-b_{n})}&lt;b^2_{n}&lt;\cdots&lt;(b_{1})^{2^{n}}=\dfrac{1}{2^{2^n}}$$
where use $\ln{(1+x)}&lt;x,x&gt;-1$
so
$$\sum_{k=1}^{n}(1-a_{k})&lt;\sum_{k=1}^{n}\dfrac{1}{2^{2^{k-1}}}?$$

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?
As used by the OP: $\enspace b_n:=1-a_n$

Examples: $\enspace\displaystyle b_1=0.5\enspace , \enspace b_2=0.5-0.5\ln 2\enspace , \enspace b_3 = b_2 + (1-b_2) \ln (1-b_2) &lt;0.013$

We have: $\enspace b_n&lt;-\ln(1-b_n)\enspace$ => $\enspace a_{n+1}=a_n(1-\ln a_n)&gt;a_n(1+b_n)\enspace$ => $\enspace b_{n+1}&lt;b_n^2\enspace$ 

Using $\,b_{n+1}&lt;b_n^2\,$ we begin with $\,b_4=1-a_4&lt;0.0001=0.1^{2^2}\,$ and get 
$\enspace b_n&lt;0.1^{2^{n-2}}\,$ .

It follows:

$\displaystyle \sum\limits_{k=1}^n (1-a_k) &lt; b_1 + b_2 + b_3 + \sum\limits_{k=4}^\infty 0.1^{2^{k-2}}$ 

$\displaystyle &lt; 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} $

$\displaystyle &lt; 1- 0.5 \ln 2 + 0.013 + 0.00010002 &lt; 0,66652643 &lt;\frac{2}{3}$
You can add more coefficients in the Taylor expansion
$$\log(1-x)\leq-x-x^2/2+...$$ 
in order to obtain the sharper bound
$$b_{n}+(1-b_{n})\ln{(1-b_{n})}&lt;b^2_{n}/2+b_n^3/2$$
and so on, if it is necessary.