Let f(x) be the generating function of the series $a_n$ ، $n\geq0$.

Question

How can I express the generatiing function of $\frac{1}{n}a_n$, $n\geq1$? I tried with integral but something went wrong with me.

Answer 1

Suppose $f$ is the generating function of $a_n$, and $g$ is the generating function of $\frac{a_n}{n}$. Then $$g’(x) = \sum_{n = 1}^{\infty} (\frac{a_n}{n}x^n)’ = \sum_{n = 1}^{\infty} (a_n x^{n-1}) = \frac{f(x) - a_0}{x}$$ Thus, by Newton-Leibniz formula $$g(x) = \int_0^x \frac{f(t)-a_0}{t}dt$$