Question 1032759
{{{ log((1+x))+log((1+x^2))+log((1+x^4)) + log((1+x^8)) }}}+...

=log({{{(1+x)(1+x^2)(1+x^4)(1+x^8)}}}*...)

=log({{{1+x + x^2 + x^3 + x^4 + x^5)}}}+...)

={{{log((1/(1-x))) = -log((1-x))}}}.

The infinite geometric series {{{1+x + x^2 + x^3 + x^4 + x^5)}}}+... converges to {{{1/(1-x)}}} by virtue of {{{0 < x < 1}}}.