How do you find the sum of #ln ( 1 + 1 / (2^(2^n)) )# from n=0 to infinity? Thanks!?

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Feb 9, 2018

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# \ #

# \qquad \qquad \qquad \qquad \quad \quad \ lim_{n rarr \infty} sum_{k=0}^{n} ln( 1 + 1/{ 2^{2^k} } ) \ = \ ln(2). #

Explanation:

# \ #

# " 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} } ].#

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