show that $\sum_{k=1}^{n}(1-a_{k})<\frac{2}{3}$

Solution 1:

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} $

$\displaystyle < 1- 0.5 \ln 2 + 0.013 + 0.00010002 < 0,66652643 <\frac{2}{3}$