# [seqfan] Re: Logarithm related identities (generated by me)

Jaume Oliver i Lafont joliverlafont at gmail.com
Sat Dec 27 02:01:49 CET 2008

```Hello Alexander and Seqfans

>> ln(3) = 1/4*(1+ Sum((1/(9)^(k+1))*(27/(2*k+1)+4/(2*k+2)+1/(2*k+3)), k = 0 ..infinity) )

This is a particular case of the more general
ln(3)=(1/(a+b))(b+Sum((1/(9)^(k+1))*(9a/(2k+1)+(a+b)/(2k+2)+b/(2k+3)))
, when a=3 and b=1 are used.
This one is obtained by summing
a*ln(3)=Sum((1/(9)^(k+1))*(a/(2k+1)+a/(2k+2))
and
b*ln(3)=b+Sum((1/(9)^(k+1))*(b/(2k+2)+b/(2k+3))

Therefore, an infinite number of similar formulas can be easily
obtained. For example, a and b can be chosen in such a way that
b/(a+b) is close to ln(3). The only requirement is that a+b doesn't
cancel (so we cannot get rid of the term in (2k+2), unless we
substitute it by another one with an even offset). If it does cancel,
we obtain another formula for zero.

>>ln(2) = 1/4*(3 - sum(1/(n*(n+1)*(2*n+1)), n=1...infinity))

This one is published at
http://numbers.computation.free.fr/Constants/Log2/log2Formulas.html
When comparing it to the next one at the list, it may be interesting
to rewrite it as

ln(2)=3/4-sum(1/((2n)(2n+1)(2n+2)), n=1..infinity)
or even
ln(2)=3/4-sum((2n-1)!/(2n+2)!,n=1..infinity)

>> ln(2) = 105*(319/44100 - sum(1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)),n=1...infinity) )
1/((2n)(2n+1)(2n+3)(2n+5)(2n+7)) = 8(n+3)!(2n-1)!/(n!(2n+7)!)

>> ln(2) = (319/420 - 3/2*sum(1/(6*n^2+39*n+63),n=1...infinity))

The last one is also similar to the two previous formulas, since
6n^2+39^n+63=(3/2)(2n+6)(2n+7)
Therefore, a simpler version is
ln(2)=  319/420 - sum(1/((2n+6)(2n+7)),n=1..infinity)
or
ln(2)= 319/420 - sum((2n+5)!/(2n+7)!, n=1..infinity)

Best wishes,
Jaume

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