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Consider the following sum for positive integers $a$ and $b$

$$\sum_{n = 0}^\infty \frac{(-1)^n}{an + b}. $$

What is the closed form of this sum? Solutions, partial progress, and suggestions appreciated.

Note: The original problem asks for $a = 8, b = 3$, though we haven't solved this case yet.

Things my friends and I tried: For the case $a = 8, b = 3$, it is equivalent to $\int_0^1 \frac{x^2}{1 + x^8} \, dx$, not much progress here. Next, we tried writing $f(x) = \sum \frac{x^n}{an+b}$ or $f(x) = \sum \frac{\cos(nx)}{an + b}$ and writing differential equations.

Alan Yan
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  • What if $a=1$, $b$ - integer. The best we can do is to subtract $b$ members from full series – Sergey Zaitsev Jun 27 '18 at 04:54
  • If I'm not mistaken, you can evaluate the integral by using complex numbers. I seem to remember working this out for integer $a$ and $b$ years ago. I don't recall whether I got a closed-form formula, or just a method for handling specific cases. The denominator splits into linear factors, and you use partial fractions. You may have to be careful about which branch of the logarithm you're on. I remember considering this, but not what I decided. – saulspatz Jun 27 '18 at 05:07

2 Answers2

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$$\sum_{n = 0}^\infty \frac{(-1)^n}{a\,n + b}=\frac 1 a\,\Phi \left(-1,1,\frac{b}{a}\right)$$ where appears the Lerch transcendent function.

For $a=8$ and $b=3$, the result expresses in terms of the digamma function; it is $$\frac{1}{16} \left(\psi ^{(0)}\left(\frac{11}{16}\right)-\psi ^{(0)}\left(\frac{3}{16}\right)\right)\approx 0.273898$$ For $$I=\int_0^1 \frac{x^2}{1 + x^8} \, dx$$ $$I==\frac{1}{8} \left(\cos \left(\frac{\pi }{8}\right) \left(\pi -2 \tanh ^{-1}\left(\sin \left(\frac{\pi }{8}\right)\right)\right)-\sin \left(\frac{\pi }{8}\right) \left(\pi -2 \tanh ^{-1}\left(\cos \left(\frac{\pi }{8}\right)\right)\right)\right)$$

Claude Leibovici
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The sum, in general, can be written as a difference of digamma values at rational arguments: \begin{align}\sum_{n\ge0}\frac{(-1)^n}{an+b}&=\sum_{n\ge0}\bigg(\frac1{2an+b}-\frac1{a(2n+1)+b}\bigg)\\[2ex] &=\sum_{n\ge0}\bigg(\frac1{2an+b}-\frac1{2an}+\frac1{2an}-\frac1{2an+a+b}\bigg)\\[2ex] &=\frac1{2a}\sum_{n\ge0}\bigg(\frac1{n+b/(2a)}-\frac1n-\frac1{n+1/2+b/(2a)}+\frac1n\bigg)\\[2ex] &=\frac1{2a}\Big[\gamma+\psi\Big(\frac32+\frac b{2a}\Big)-\gamma-\psi\Big(1+\frac b{2a}\Big)\Big] \end{align}and Gauss's digamma theorem provides the formula to evaluate them (apart from possible trig simplifications, it doesn't get any cleaner).

Antonio DJC
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