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?

  • $\begingroup$ This question is not very clear-at least to me. $\endgroup$ – picakhu Nov 23 '11 at 17:01
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    $\begingroup$ Why you don't just replace $xz$ by (say) $t$ so that the question is simpler? (this does not answer it, mind you). $\endgroup$ – leonbloy Nov 23 '11 at 17:02
  • $\begingroup$ Do you know the answer when $\alpha = 0$? $\endgroup$ – Qiaochu Yuan Nov 23 '11 at 17:06
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    $\begingroup$ Also, what are $\alpha$ and $\beta$? $\alpha$ had better be nonzero, if you're going to avoid a zero denominator in the first term. $\endgroup$ – Dimitrije Kostic Nov 23 '11 at 17:06
  • $\begingroup$ @picakhu: i just want to find the suitable equation for related summation. $\endgroup$ – DRN Nov 23 '11 at 17:06

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: $$ f(w) = \sum_{n=0}^\infty \frac{w^n}{\beta n +\alpha} = \sum_{n=0}^\infty w^n \int_0^1 u^{n \beta + \alpha-1} \mathrm{d} u = \int_0^1 \frac{u^{\alpha-1}}{1-w u^\beta} \mathrm{d} u $$

  • $\begingroup$ The computation of $c$ is odd. All one knows is that for every $(\alpha,\beta)$ there is some $c_{\alpha,\beta}$ such that $f(w)$ is so and so, and you show that $c_{\alpha,\alpha}=0$ for every $\alpha$. Right. But this seems to give no information at all on $c_{\alpha,\beta}$ for $\alpha\ne\beta$. $\endgroup$ – Did Nov 24 '11 at 7:41
  • $\begingroup$ @DidierPiau Yes, but I can justify $c_{\alpha,\beta} = 0$ a-posteriori, by rexpanding the integrand. I have expanded the answer. Of course, the integral representations in my post work for $\frac{\alpha}{\beta}>0$. $\endgroup$ – Sasha Nov 24 '11 at 16:02
  • $\begingroup$ thanks all.. any other ways to find the expression instead of using Lerch zeta function? $\endgroup$ – DRN Nov 25 '11 at 6:34
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    $\begingroup$ @Norlyda Not for generic $\alpha$, $\beta$. Notice, though, that for $\alpha = \beta/2$, $f(w) = \frac{2}{\beta \sqrt{w}} \arctan(\sqrt{w})$. $\endgroup$ – Sasha Nov 25 '11 at 6:40
  • $\begingroup$ @Sasha: my task is actually to form that summation in form of this expression, $\sum_{n=0}^{\infty}{x^n}{z^n}=\frac{1}{1-xz}$, maybe it is impossible,right? $\endgroup$ – DRN Nov 25 '11 at 7:35

$$\sum\limits_{n=0}^{+\infty}\frac{x^n}{n\beta+\alpha}= \frac1{\beta x^{\alpha/\beta}}\int\limits_0^{x}\frac{u^{\alpha/\beta}}{1-u}\frac{\mathrm du}u $$


$$\sum_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$$

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!


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