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?

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.
