show that `sum_2^(k->oo)ln(1-1/k^2)=-ln2`

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Show `sum_2^(k->oo)ln(1-1/k^2)=-ln2` :

Here is one approach:

Write out the terms of the series:

`ln(1-1/4)+ln(1-1/9)+ln(1-1/16)+ln(1-1/25)+` ...

`=ln(3/4)+ln(8/9)+ln(15/16)+ln(24/25)+` ...

`=ln(3/4*8/9*15/16*24/25*...(n^2-1)/n^2*...)`

The infinite product can be represented by `prod_2^(oo) (n^2-1)/n^2`

or `prod_2^(oo) 1-1/n^2` which is 1/2. (See number 21 in Mathworld link)

Thus `ln(3/4*8/9*15/16*24/25*...)=ln(1/2)`

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`sum_2^(k->oo)ln(1-1/k^2)=ln(prod_2^(oo)(1-1/n^2))=ln(1/2)=-ln2`

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Sources:

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