What is the Generating Function for the sequence of reciprocal triangular numbers?

Question

Starting from $$0+1x+\frac{1}{2}x^2+\frac{1}{3}x^3+\frac{1}{4}x^4+\dots : GF=-\ln(1-x)$$ I can get to $$1+\frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{4}x^3+\frac{1}{5}x^4\dots : GF=-\frac{ln(1-x)}{x}$$ and I'd now like to subtract $$\frac{1}{2}+\frac{1}{3}x+\frac{1}{4}x^2+\frac{1}{5}x^3+\frac{1}{6}x^4\dots : GF=?$$ with a view to then doubling everything to get the generating function for $$1+\frac{1}{3}x+\frac{1}{6}x^2+\frac{1}{10}x^3+\frac{1}{15}x^4+\frac{1}{21}x^5\dots : GF=?$$ I feel this should be easy but can't seem to get it exactly

(If I right shift the $1+\frac{1}{2}x+\frac{1}{3}x^2+\dots$ I'll be subtracting an unwanted $2$.)

Answer 1

First, you made a mistake in the second series: $$1+\frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{4}x^3+\frac{1}{5}x^4\dots : GF=-\frac{\ln(1-x)}{x}.$$ Then $$\frac{1}{2}+\frac{1}{3}x+\frac{1}{4}x^2+\frac{1}{5}x^3+\frac{1}{6}x^4\dots : GF=\frac{\frac{-\ln(1-x)}{x}-1}{x} = -\frac{x+\ln(1-x)}{x^2}.$$ (That is, subtract $1$ from the first series and divide by $x$.) Subtracting the two and multiply by two, giving $$1+\frac{1}{3}x+\frac{1}{6}x^2+\frac{1}{10}x^3+\frac{1}{15}x^4+\frac{1}{21}x^5\dots : GF=\frac{2(x+\ln(1-x)-x\ln(1-x))}{x^2}.$$

Answer 2

When you substract first and second, you get $$-1+\frac{1}{2}x+\frac{1}{6}x^2+\frac{1}{12}x^3+\frac{1}{20}x^4\dots=f(x)-g(x)$$ Then multiplying by 2, you get $$-2+x+\frac{1}{3}x^2+\frac{1}{6}x^3+\frac{1}{10}x^4+\dots=2(f(x)-g(x))$$ So, $$x+\frac{1}{3}x^2+\frac{1}{6}x^3+\frac{1}{10}x^4+\dots=2(f(x)-g(x))+2$$ Finally, dividing by $x$. $$1+\frac{1}{3}x+\frac{1}{6}x^2+\frac{1}{10}x^3+\dots=\frac{2(f(x)-g(x))+2}{x}$$