Compute $\sum\limits_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$
Known that $\sum_{n=0}^{\infty}{x^n}{z^n}=\frac{1}{1-xz}$. If we have $\sum_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$ where $\beta, \alpha $ are element of real numbers but not equal $0$. What is a suitable expression for that summation?
Solution 1:
$$\sum\limits_{n=0}^{+\infty}\frac{x^n}{n\beta+\alpha}= \frac1{\beta x^{\alpha/\beta}}\int\limits_0^{x}\frac{u^{\alpha/\beta}}{1-u}\frac{\mathrm du}u $$
Solution 2:
Since $x$ and $z$ only occur as a product, denote $w = x z$. We seek to evaluate $ f(w) = \sum_{n=0}^\infty \frac{w^n}{\beta n + \alpha}$.
Notice that $\frac{w^n}{n \beta + \alpha} = \frac{1}{\beta} w^{-\frac{\alpha}{\beta}} \int w^{n - 1+ \frac{\alpha}{\beta}} \mathrm{d} w$. Therefore $$ f(w) = \frac{1}{\beta} w^{-\frac{\alpha}{\beta}} \int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z + \frac{c}{\beta} w^{-\frac{\alpha}{\beta}} $$ The constant $c$ is determined by setting $\alpha = \beta$. Then, using the series, $f(w) = \frac{1}{\beta w} \sum_{n=0}^\infty \frac{w^{n+1}}{n+1} = -\frac{1}{\beta w} \log(1-w)$.
So we get: $$ -\frac{1}{\beta w} \log(1-w) = \frac{1}{\beta w} \int_0^w \frac{1}{1-z} \mathrm{d} z + \frac{c}{\beta w} $$ From where it follows that $c=0$. Therefore the sum admits the following integral representation: $$ f(w) = \frac{1}{\beta} w^{-\frac{\alpha}{\beta}} \int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z = \frac{1}{\beta} \int_0^1 \frac{u^{\frac{\alpha}{\beta}-1}}{1-w u} \mathrm{d} u = \int_0^1 \frac{u^{\alpha-1}}{1- w u^\beta} \mathrm{d} u $$ This integral define a special function, known as Lerch zeta function.
Actually the question from Didier Piau, made me realize that I missed a simpler route: $$ f(w) = \sum_{n=0}^\infty \frac{w^n}{\beta n +\alpha} = \sum_{n=0}^\infty w^n \int_0^1 u^{n \beta + \alpha-1} \mathrm{d} u = \int_0^1 \frac{u^{\alpha-1}}{1-w u^\beta} \mathrm{d} u $$
Solution 3:
$$\sum_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$$
is expressible in terms of the Lerch transcendent:
$$\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s}$$
Taking $s=1$ gives
$$\Phi(z,1,a)=\sum_{k=0}^\infty \frac{z^k}{a+k}$$
after which
$$\begin{align*}\Phi(xz,1,a)&=\sum_{k=0}^\infty \frac{(xz)^k}{a+k}\\ \Phi(xz,1,\alpha/\beta)&=\sum_{k=0}^\infty \frac{(xz)^k}{\alpha/\beta+k}\\ \frac1{\beta}\Phi(xz,1,\alpha/\beta)&=\sum_{k=0}^\infty \frac{(xz)^k}{\alpha+\beta k}\end{align*}$$
and presto!