The log series distribution is named because of its similarity to the logarithmic series:
 x^{1}/1 + x^{2}/2 − x^{3}/3 + x^{4}/4 − x^{5}/5 + x^{6}/6... = Σ[(1)^{(i1)}x^{i}/i]
If −1<x<1 this series is infinitely long and convergent. Its sum is log_{e}(1+x)  hence its name.

If all the terms in the series are positive, then x^{1}/1 + x^{2}/2 + x^{3}/3 + x^{4}/4 + ax^{5}/5 + x^{6}/6... = Σ[x^{i}/i]
Provided that 0<x<1 the sum of this series is −log_{e}(1−x).
This distribution can be conveniently rescaled by multiplying each term by a constant, α.
Its sum is Σ[αx^{i}/i] = αΣ[x^{i}/i] = −αlog_{e}(1−x).
If α = ^{1}/−log_{e}(1−x) then the sum of this series is one. This is the mass probability function of the log series distribution  as shown graphically below.
{Fig. 1}
Notice that this relationship is true irrespective of the value of x.
Provided this model describes the number of organisms (i) observed within each of s species, to work out the number of species represented by i individuals we set α to ^{s}/−log_{e}(1−x). In other words we multiply the height of each bar in the graph above by s.
 If we multiply each term in the above series by i, we obtain a simpler series: x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}... = Σ[x^{i}] and, provided that 0<x<1 then its sum (n) is x/[1  x].
Once again, if we rescale these terms by multiplying each by a constant, α, their sum is αx/[1  x].
So if each bar on the graph above corresponds to a set of species, each represented by i organisms, the total number of organisms in each bar must be s×i. The total number of organisms, n, is
αx/[1  x]  and therefore
α = n[1  x]/x.