I'm trying to find the $[x^n]$ from a sequence below:
$$1, \frac{3}{2}, \frac{7}{3}, \frac{13}{4}, \frac{21}{5}, \frac{31}{6}, ...$$
I figured out the following sum generates that series:
$$R(x) = \sum_{k=0}^n \frac{(k+1)^2 - k}{k+1} . x^k$$
Rearranging the sum:
$$R(x) = \sum_{k=0}^n \frac{(k+1)^2}{k+1} . x^k - \sum_{k=0}^n \frac{k}{k+1} . x^k$$
$$R(x) = \sum_{k=0}^n (k+1) . x^k - \sum_{k=0}^n \frac{k}{k+1} . x^k$$
Let $F(x)$ and $H(x)$ to be the sums above respectively:
$$R(x) = F(x) - H(x)$$
Considering a known generating function $L(x)$:
$$L(x) = \ln \left(\frac{1}{1-x}\right)$$
and by making some manipulations I get:
$$F(x) = \frac{1}{(1-x)^2}$$
$$H(x) = \dfrac{d}{dx} \left(\frac{L(x)}{x}\right)x $$
Calculating I get:
$$H(x) = \frac{1}{1-x} - \frac{\ln \left(\frac{1}{1-x}\right)}{x}$$
To finish we have:
$$R(x) = \frac{1}{(1-x)^2} + \frac{1}{1-x} - \frac{\ln \left(\frac{1}{1-x}\right)}{x}$$
I've skipped some steps to save space.
It seems to me the answer is not correct. I was wondering at what point I'm failing.. Does anyone here can help me?
Just a sign error on the last row: $R=F\color{red}{-}H$.
Note also that $\log(\frac1{1-x})=-\log(1-x)$, slightly simpler.