What is the value of $\sum_{n=0}^\infty\frac{(-1)^n}{an^2+bn+c}$

Asked Jul 01 '23 at 16:51

Active Jul 01 '23 at 16:51

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I want to evaluate the following alternating series: $$\sum_{n=0}^\infty\frac{(-1)^n}{an^2+bn+c}$$ I believe a closed form does exist, since there is one for this series.

I thought that one approach might be to factor the denominator in its two roots, and then rewrite using partial fractions. Now the two series obtained are of the form of this series, and so there is a closed form, but the calculations get extremely messy after a few steps.

So I wonder, is there another method to get to a more pleasing closed form? Like the one in the first link?

asked Jul 01 '23 at 16:51

Zima

2,925

partial sum – infinitezero Jul 01 '23 at 17:17
2

You can apply your first link to the two series ($n$ even, $n$ odd) of which yours is the difference.$$\sum_{n=0}^\infty\frac{(-1)^n}{an^2+bn+c}=\sum_{k=0}^\infty\frac1{(4a)k^2+(2b)k+c}-\sum_{k=0}^\infty\frac1{(4a)k^2+(4a+2b)k+c}$$ – Anne Bauval Jul 01 '23 at 17:57

What is the value of $\sum_{n=0}^\infty\frac{(-1)^n}{an^2+bn+c}$

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