# \ #
# " Let's start with the observation:" #
# ( 1 - 1/{ 2^{2^n} } ) ( 1 + 1/{ 2^{2^n} } ) \ = \ ( 1 - ( 1/{ 2^{2^n} } )^2 ) #
# \qquad \qquad = \ 1 - 1/{ 2^{ 2 \cdot 2^n} } \ = \ 1 - 1/{ 2^{ 2^{n+1} } } #
# " Thus:" #
# ( 1 - 1/{ 2^{2^n} } ) ( 1 + 1/{ 2^{2^n} } ) \ = \ 1 - 1/{ 2^{ 2^{n+1} } } \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1) #
# " So, iterating this:" #
# ( 1 - 1/{ 2^{2^0} } ) ( 1 + 1/{ 2^{2^0} } )( 1 + 1/{ 2^{2^1} } ) ( 1 + 1/{ 2^{2^2} } ) \cdots ( 1 + 1/{ 2^{2^n} } ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1 - 1/{ 2^{ 2^{n+1} } }. #
# \ #
# "I.e.:" \qquad \qquad ( 1 - 1/{ 2^{2^0} } ) prod_{k=0}^{n} ( 1 + 1/{ 2^{2^k} } ) \ = \ 1 - 1/{ 2^{ 2^{n+1} } }. \qquad \qquad \qquad (2)#
# "Taking logarithm of both sides of this equation:" #
# \qquad \qquad ln [ ( 1 - 1/{ 2^{2^0} } ) prod_{k=0}^{n} ( 1 + 1/{ 2^{2^k} } ) ] \ = \ ln [1 - 1/{ 2^{ 2^{n+1} } } ].#
# :. \ ln [ ( 1 - 1/{ 2^{2^0} } ) ] + [ sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) ] \ = \ ln [1 - 1/{ 2^{ 2^{n+1} } } ].#
# :. \ sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) \ = \ ln [1 - 1/{ 2^{ 2^{n+1} } } ] - ln [ 1 - 1/{ 2^{2^0} } ].#
# :. \ lim_{n rarr \infty} sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) #
# \qquad \qquad \qquad \qquad = \ lim_{n rarr \infty}[ ln [1 - 1/{ 2^{ 2^{n+1} } } ] - ln [ 1 - 1/{ 2^{2^0} } ] ]. #
# :. \quad \quad \ lim_{n rarr \infty} sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) \ = \ ln [1 - 0 ] - ln [ 1 - 1/{ 2 } ]. #
# :. \quad \quad \ lim_{n rarr \infty} sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) \ = \ ln(1) - ln ( 1/{ 2 } ) #
# \qquad \qquad \qquad \qquad = \ 0 - ( -ln(2 ) )\ = \ ln(2). #
# \ #
# "So now we have determined the desired limit." #
# "Statement of result:" #
# \qquad \qquad \qquad \qquad \quad \quad \ lim_{n rarr \infty} sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) \ = \ ln(2). \qquad \quad square #
# \ #
# \ #
# "Remark:" #
# "In the above, we also established a perhaps more subtle result --" #
# "we have found closed a formula for the finite version of this sum:" #
# :. \ sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) \ = \ ln [1 - 1/{ 2^{ 2^{n+1} } } ] - ln [ 1 - 1/{ 2^{2^0} } ].#