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given the series

$$ \sum_{n=0}^{\infty} \frac{x^{n}}{an+b}=f(x) $$

so for any postive integer $ an+b $ is never 0

how could i get $ f(x) $ from this Taylor series ??

Jose Garcia
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  • (See the topic linked) Lerch function it is, but there is no real simplification for exponent $s=1$, it has integral representation as well. For $x=-1$ it can be expressed with diagamma https://math.stackexchange.com/questions/2833264/generalization-of-alternating-harmonic-series – zwim Jan 21 '20 at 21:34

1 Answers1

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For $|x|<1$ and $a \ne 0$, this is $$ {\frac {1}{a}{\Phi} \left( x,1,{\frac {b}{a}} \right) }$$ Where $\Phi$ is the Lerch Phi function or Lerch transcendent.

There are some special cases, e.g.: $$ \Phi(x,1,1/2) = \frac{1}{\sqrt{z}} \ln \left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)$$

EDIT:

More generally, if $a$ and $b$ are integers with $0 < b/a < 1$, and $\omega = \exp(2 \pi i/a)$, for $|x| < 1$ we have $$ \sum_{j=1}^\infty \frac{\omega^{kj} x^j}{j} = - \ln(1 - \omega^k x)$$ so that $$-\sum_{k=0}^{a-1} \omega^{-kb} \ln(1-\omega^k x) = a \sum_{j \equiv b \mod a} \frac{\omega^{kj} x^j}{j} = x^b \Phi(x^a,1,b/a) $$ i.e $$ \Phi(z,1,b/a) = - z^{-b/a} \sum_{k=0}^{a-1} \exp(-2\pi i kb/a) \ln(1 - \exp(2\pi i k/a) z^{1/a})$$

Robert Israel
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