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?
Compute $\sum\limits_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$
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0This question is not very clear-at least to me. – 2011-11-23
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2Why you don't just replace $xz$ by (say) $t$ so that the question is simpler? (this does not answer it, mind you). – 2011-11-23
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0Do you know the answer when $\alpha = 0$? – 2011-11-23
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1Also, what are $\alpha$ and $\beta$? $\alpha$ had better be nonzero, if you're going to avoid a zero denominator in the first term. – 2011-11-23
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0@picakhu: i just want to find the suitable equation for related summation. – 2011-11-23
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0Oh, you mean a *solution/expression*? Thanks, that helps. – 2011-11-23
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0@leonbloy: yes, it doesn't matter if use $t$ or $xz$ – 2011-11-23
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0@QiaochuYuan: $\alpha, \beta $ just a constant value – 2011-11-23
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0@DimitrijeKostic: yes, $\alpha, \beta \neq 0$ – 2011-11-23
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0It is not trivial. [wolfram](http://www.wolframalpha.com/input/?i=sum+from+n%3D0+to+infty+(t^n%2F%28n*A%2BB%29)) – 2011-11-23
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0I dont know why the link doesnt work. Copy and paste this. http://www.wolframalpha.com/input/?i=sum+from+n%3D0+to+infty+{t^n+%2F+%28nA%2BB%29} – 2011-11-23
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0@picakhu: thanks for introducing that software.. – 2011-11-23
3 Answers
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: $$ f(w) = \sum_{n=0}^\infty \frac{w^n}{\beta n +\alpha} = \sum_{n=0}^\infty w^n \int_0^1 u^{n \beta + \alpha-1} \mathrm{d} u = \int_0^1 \frac{u^{\alpha-1}}{1-w u^\beta} \mathrm{d} u $$
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0The 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 $\alpha\ne\beta$. – 2011-11-24
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0@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 $\frac{\alpha}{\beta}>0$. – 2011-11-24
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0thanks all.. any other ways to find the expression instead of using Lerch zeta function? – 2011-11-25
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1@Norlyda Not for generic $\alpha$, $\beta$. Notice, though, that for $\alpha = \beta/2$, $f(w) = \frac{2}{\beta \sqrt{w}} \arctan(\sqrt{w})$. – 2011-11-25
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0@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? – 2011-11-25
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0The route to $c=0$ still seems wrong, except when $\alpha=\beta$. – 2011-11-25
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0@Sasha: please help me to solve [this](http://math.stackexchange.com/questions/87775/another-problem-of-solving-summation) too.. really need your guidance. Tqsm. – 2011-12-03
$$\sum\limits_{n=0}^{+\infty}\frac{x^n}{n\beta+\alpha}= \frac1{\beta x^{\alpha/\beta}}\int\limits_0^{x}\frac{u^{\alpha/\beta}}{1-u}\frac{\mathrm du}u $$
$$\sum_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$$
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!